9231 Differential equations

9231-2021-mj-21-q02

May/June 2021 · Paper 21 · Question 2 · 7 marks

2(a) m2 +3m+2=0 leading to m = -1, -2 M1 Auxiliary equation. y= Ae −x +Be −2x A1 Complimentary function. y= p+qx leading to y'=q leading to y''=0 B1 Particular integral and its derivatives. 3q+2(p+qx)=2x+1 M1 Substitutes and equates coefficients. q=1 p=−1 A1 y= Ae −x +Be −2x +x−1 A1 Must see ‘y =’. 6 © UCLES 2021 Page 6 of 13 2(b) y = x −1 B1 FT Accept y ≈ x −1. 1 Question Answer Marks Guidance

Official mark scheme pages: 6, 7 · source PDF URL

9231-2021-mj-21-q04

May/June 2021 · Paper 21 · Question 4 · 9 marks

4 dy tan1θ B1 Divides through by sinθ. +cosecθy = 2 dθ sinθ e cosecθdθ =elntan1 2 θ= tan 1θ M1 A1 Finds integrating factor. 2 d ( ) tan2 1θ  1sin1θ M1 Correct form on LHS and uses an appropriate ytan1θ = 2 = 1tan1θsec2 1θ  = 2 2  identity. dθ 2 sinθ 2 2 2  cos3 1 2 θ  ( ) ytan1θ= 1tan2 1θ+C or 1sec2 1θ+C M1 A1 Integrates RHS. 2 2 2 2 2 1 M1 Substitutes initial conditions. 1= +C 2 y=1 ( tan1θ+cot1θ )[ =cosecθ ] M1 A1 Divides through by their integrating factor. 2 2 2 9 © UCLES 2021 Page 8 of 13

Official mark scheme pages: 8 · source PDF URL

9231-2021-mj-22-q02

May/June 2021 · Paper 22 · Question 2 · 7 marks

2(a) m2 +3m+2=0 leading to m = -1, -2 M1 Auxiliary equation. y= Ae −x +Be −2x A1 Complimentary function. y= p+qx leading to y'=q leading to y''=0 B1 Particular integral and its derivatives. 3q+2(p+qx)=2x+1 M1 Substitutes and equates coefficients. q=1 p=−1 A1 y= Ae −x +Be −2x +x−1 A1 Must see ‘y =’. 6 © UCLES 2021 Page 6 of 13 2(b) y = x −1 B1 FT Accept y ≈ x −1. 1 Question Answer Marks Guidance

Official mark scheme pages: 6, 7 · source PDF URL

9231-2021-mj-22-q04

May/June 2021 · Paper 22 · Question 4 · 9 marks

4 dy tan1θ B1 Divides through by sinθ. +cosecθy = 2 dθ sinθ e cosecθdθ =elntan1 2 θ= tan 1θ M1 A1 Finds integrating factor. 2 d ( ) tan2 1θ  1sin1θ M1 Correct form on LHS and uses an appropriate ytan1θ = 2 = 1tan1θsec2 1θ  = 2 2  identity. dθ 2 sinθ 2 2 2  cos3 1 2 θ  ( ) ytan1θ= 1tan2 1θ+C or 1sec2 1θ+C M1 A1 Integrates RHS. 2 2 2 2 2 1 M1 Substitutes initial conditions. 1= +C 2 y=1 ( tan1θ+cot1θ )[ =cosecθ ] M1 A1 Divides through by their integrating factor. 2 2 2 9 © UCLES 2021 Page 8 of 13

Official mark scheme pages: 8 · source PDF URL

9231-2021-mj-23-q05

May/June 2021 · Paper 23 · Question 5 · 10 marks

5(a) y=kxe −x leading to y'=ke −x ( 1−x ) leading to y''=ke −x ( x−2 ) B1 B1 Differentiates particular integral. ke −x ( x−2 )−2ke −x ( 1−x )−3kxe −x =4e −x leading to −4k =4 M1 Substitutes. k =−1 A1 4 5(b) m2 −2m−3=0 leading to m=−1,3 M1 Auxiliary equation. y= Ae −x +Be3x A1 Complementary function. (Can be awarded if seen in the general solution.) y= Ae −x +Be3x −xe −x =(A−x)e −x +Be3x A1 FT General solution. FT on their k. Must have “y=”. y'=−(A−x)e −x −e −x +3Be3x B1 Derivative of general solution. A+B= 1, −A−1+3B= 1 leading to A=0, B= 1 M1 Substitutes initial conditions into general solution. 2 2 2 y= 1e3x −xe −x A1 Must have “y=”. 2 6 © UCLES 2021 Page 10 of 15

Official mark scheme pages: 10 · source PDF URL

9231-2021-mj-23-q06

May/June 2021 · Paper 23 · Question 6 · 10 marks

6(a) sinhx= 1 ex −e −x coshx= 1 ex +e −x B1 Writes in exponential from. 2 2 1 ( ex −e −x )2 = 1 ( e2x −2+e −2x ) = 1 ( e2x +e −2x ) −1 M1 Expands. 2 2 2 cosh2x−1 A1 AG. 3 6(b) e cothx =elnsinhx =sinhx M1 A1 Finds integrating factor. d M1 Correct form on both sides. (ysinhx)=4sinh2x dx ( ) ysinhx=2 1sinh2x−x +C=sinh2x−2x+C M1 A1 Integrates sinh2xusing identity. 2 4 40 M1 Substitutes initial conditions into their solution. sinhln3=sinhln9−2ln3+C leading to = −2ln3+C 3 9 28 A1 ysinhx=sinh2x−2x+2ln3− 9 7 © UCLES 2021 Page 11 of 15

Official mark scheme pages: 11 · source PDF URL

9231-2021-on-21-q05

Oct/Nov 2021 · Paper 21 · Question 5 · 11 marks

5 m2 −2m+1=0m=1 M1 Auxiliary equation. y=ex ( Ax+B ) A1 Complementary function. Accept ‘y =’ missing. y= psinx+qcosx, y'= pcosx−qsinx, y''=−psinx−qcosx M1 A1 Particular integral and its derivatives. −p+2q+ p=0 −q−2p+q=4 M1 Substitutes and equates coefficients. p=−2 q=0 A1 y=ex ( Ax+B )−2sinx A1 y'= Aex +ex ( Ax+B )−2cosx M1 Differentiates. B=−4 A+B−2=3 A=9 M1 A1 Forms simultaneous equations using initial conditions. y=ex ( 9x−4 )−2sinx A1 11 © UCLES 2021 Page 9 of 13

Official mark scheme pages: 9 · source PDF URL

9231-2021-on-21-q07

Oct/Nov 2021 · Paper 21 · Question 7 · 11 marks

7(a) dy 1 x2 −x x2 −1 B1 Divides through by x2−1. + y= dx x2 −1 x2 −1 1 M1 A1 Finds integrating factor. M1 for correct form  dx e √(x2−1) =ecosh−1x 1 a dx e √(x2−1) , where a is a non-zero constant. x+ x2 −1 A1 AG 4 ( ( )) 7(b) d x M1 A1 Correct form on LHS and simplifies RHS. y x+ x2 −1 = dx x2 −1 ( ) M1 A1 Integrates RHS. For M1, needs to integrate non-zero y x+ x2 −1 = x2 −1+C x multiple of . x2 −1 2= 3+C leading toC= 5 M1 Substitutes initial conditions. 4 4 x2 −1+ 5 M1 A1 Divides through by coefficient of y. y= 4 x+ x2 −1 7 © UCLES 2021 Page 11 of 13

Official mark scheme pages: 11 · source PDF URL

9231-2021-on-22-q02

Oct/Nov 2021 · Paper 22 · Question 2 · 7 marks

2 e ln x4+5 =x4 +5 M1 A1 Finds integrating factor. d ( ( )) M1 Writes in correct form. y x4 +5 =6x5 +30x dx y(x4 +5)=x6 +15x2 +C A1 6=16+C M1 Substitutes initial conditions. x6 +15x2 −10 M1 A1 Division through by coefficient of y. y= x4 +5 7 Question Answer Marks Guidance

Official mark scheme pages: 7 · source PDF URL

9231-2021-on-22-q07

Oct/Nov 2021 · Paper 22 · Question 7 · 11 marks

7(a) dy dw B1 dw 1 dy 2 =x2 +2xw = − y dx dx dx x2 dx x3 d2y d2w dw B1 d2w 1 d2y 4 dy 6 =x2 +4x +2w = − + y dx2 dx2 dx dx2 x2 dx2 x3 dx x4 d2y dy d2w dw dw M1 Uses substitution to find w-x equation, AG. +4 +5y=x2 +4x +2w+4x2 +8xw+5x2w dx2 dx dx2 dx dx d2w ( )dw ( ) A1 AG 5x2 + 4x2 +4x + 5x2 +8x+2 w=5x2 +4x+2 dx2 dx 4 7(b) m2 +4m+5=0 leading tom=−2±i M1 Auxiliary equation. y=e −2x ( Acosx+Bsinx ) A1 Complimentary function. Allow “y =” missing. Don’t allow e.g. e −2x ( Acosx+Bisinx ) . y= px2 +qx+r , y'=2px+q, y''=2p B1 Particular integral and its derivatives. 2p+8px+4q+5px2 +5qx+5r=5px2 +(8p+5q)x+(2p+4q+5r) M1 Substitutes and equates coefficients. 5p=5 8p+5q=4 2p+4q+5r=2 p=1 q=−4 r = 16 A1 5 25 x2w=e −2x ( Acosx+Bsinx )+x2 −4x+16 M1 Substitutes for y and find w in terms of x. 5 25 w= ( xex )−2( Acosx+Bsinx )+1−(5x) −1+ ( 5x )−2 A1 4 4 7 © UCLES 2021 Page 12 of 14

Official mark scheme pages: 12 · source PDF URL

9231-2021-on-23-q05

Oct/Nov 2021 · Paper 23 · Question 5 · 11 marks

5 m2 −2m+1=0m=1 M1 Auxiliary equation. y=ex ( Ax+B ) A1 Complementary function. Accept ‘y =’ missing. y= psinx+qcosx, y'= pcosx−qsinx, y''=−psinx−qcosx M1 A1 Particular integral and its derivatives. −p+2q+ p=0 −q−2p+q=4 M1 Substitutes and equates coefficients. p=−2 q=0 A1 y=ex ( Ax+B )−2sinx A1 y'= Aex +ex ( Ax+B )−2cosx M1 Differentiates. B=−4 A+B−2=3 A=9 M1 A1 Forms simultaneous equations using initial conditions. y=ex ( 9x−4 )−2sinx A1 11 © UCLES 2021 Page 9 of 13

Official mark scheme pages: 9 · source PDF URL

9231-2021-on-23-q07

Oct/Nov 2021 · Paper 23 · Question 7 · 11 marks

7(a) dy 1 x2 −x x2 −1 B1 Divides through by x2−1. + y= dx x2 −1 x2 −1 1 M1 A1 Finds integrating factor. M1 for correct form  dx e √(x2−1) =ecosh−1x 1 a dx e √(x2−1) , where a is a non-zero constant. x+ x2 −1 A1 AG 4 ( ( )) 7(b) d x M1 A1 Correct form on LHS and simplifies RHS. y x+ x2 −1 = dx x2 −1 ( ) M1 A1 Integrates RHS. For M1, needs to integrate non-zero y x+ x2 −1 = x2 −1+C x multiple of . x2 −1 2= 3+C leading toC= 5 M1 Substitutes initial conditions. 4 4 x2 −1+ 5 M1 A1 Divides through by coefficient of y. y= 4 x+ x2 −1 7 © UCLES 2021 Page 11 of 13

Official mark scheme pages: 11 · source PDF URL

9231-2022-mj-21-q03

May/June 2022 · Paper 21 · Question 3 · 8 marks

3(a) z2 z10 leading to m1 i 3 M1 Auxiliary equation. 2 2 x e1 2 t  Asin 3tBcos 3t  A1 Complimentary function. 2 2 x pt2 qtr leading to x2ptq leading to x2p B1 Particular integral and its derivatives. 2p2ptq pt2 qtr t2 1 M1 Substitutes and equates coefficients. PI must have the correct form. r 1, q  2, p 1 A1 xe1 2 t Asin 3tBcos 3t  t2 2t1 A1 2 2 6 3(b) e1 2 t 0 as t  M1 x2 A1 FT 2 © UCLES 2022 Page 7 of 13

Official mark scheme pages: 7 · source PDF URL

9231-2022-mj-21-q06

May/June 2022 · Paper 21 · Question 6 · 10 marks

6 dy dv B1 vx dx dx  dv M1 Substitutes and derives first order separable equation. x  vx  vx 9x2 x2v2  dx dv A1 x  v2 9 dx 1 1 M1 Separates variables and integrates both sides.  dv dx v2 9 x v A1 sinh1   lnxC 3 C 0 B1 Substitutes initial conditions.   M1 Converts to logarithmic form. ln v  v2 1 lnx 3 9 A1 y y2 9x2 3x2 9x4 6x2y y2 y2 9x2 0 M1 Squares to eliminate radical. y  3(x2 1) A1 2 10 © UCLES 2022 Page 10 of 13

Official mark scheme pages: 10 · source PDF URL

9231-2022-mj-22-q03

May/June 2022 · Paper 22 · Question 3 · 8 marks

3(a) z2 z10 leading to m1 i 3 M1 Auxiliary equation. 2 2 x e1 2 t  Asin 3tBcos 3t  A1 Complimentary function. 2 2 x pt2 qtr leading to x2ptq leading to x2p B1 Particular integral and its derivatives. 2p2ptq pt2 qtr t2 1 M1 Substitutes and equates coefficients. PI must have the correct form. r 1, q  2, p 1 A1 xe1 2 t Asin 3tBcos 3t  t2 2t1 A1 2 2 6 3(b) e1 2 t 0 as t  M1 x2 A1 FT 2 © UCLES 2022 Page 7 of 13

Official mark scheme pages: 7 · source PDF URL

9231-2022-mj-22-q06

May/June 2022 · Paper 22 · Question 6 · 10 marks

6 dy dv B1 vx dx dx  dv M1 Substitutes and derives first order separable equation. x  vx  vx 9x2 x2v2  dx dv A1 x  v2 9 dx 1 1 M1 Separates variables and integrates both sides.  dv dx v2 9 x v A1 sinh1   lnxC 3 C 0 B1 Substitutes initial conditions.   M1 Converts to logarithmic form. ln v  v2 1 lnx 3 9 A1 y y2 9x2 3x2 9x4 6x2y y2 y2 9x2 0 M1 Squares to eliminate radical. y  3(x2 1) A1 2 10 © UCLES 2022 Page 10 of 13

Official mark scheme pages: 10 · source PDF URL

9231-2022-mj-23-q05

May/June 2022 · Paper 23 · Question 5 · 9 marks

5 dy 7y 1 B1 Divides through by x(x7).   dx x(x7) x7 7 1 1 7 7 B1 Converts to partial fractions or completes the   or  x(x7) x x7 x(x7) x 72  49 square. 2 4 x M1 A1 Finds integrating factor. Accept non-zero multiple e7x1(x7)1dx elnxln(x7)  x x7 of . x7 d  yx  x 1 7 M1 Correct form on LHS and attempt to integrate RHS. dx  x7    x72  x7  x72 RHS must be of the form a  b . x7 x72 yx A1 ln(x7)7(x7)1 (C) x7 7 ln8 7 C M1 Finds C, substitutes into their expression. 8 8  7 x7 7 M1 A1 Divides their expression through by coefficient of y. y 1  ln    x  8  x 9 © UCLES 2022 Page 10 of 15

Official mark scheme pages: 10 · source PDF URL

9231-2022-mj-23-q07

May/June 2022 · Paper 23 · Question 7 · 11 marks

7(a) 4m2 10 leading to m1, M1 Auxiliary equation. 2 y Ae1 2 x Be 1 2 x A1 Complementary function. Allow "y" missing. k 3 leading to k 3 M1 A1 Substitutes yk. y Ae1 2 x Be 1 2 x 3 A1 FT General solution. Must have y or sensible equivalent, such as y . Don’t allow GS . GS y'1 Ae1 2 x  1Be 1 2 x B1 Derivative of general solution. 2 2 AB0 1 A1B2 leading to A2,B2 M1 Substitutes initial conditions into general solution, 2 2 solves simultaneous equations. y2e 1 2 x 2e1 2 x 3 A1 Particular solution. Must have y or sensible equivalent, such as y . Don’t allow PS . PS 8 7(b) y0 leading to 4sinh1x3 leading to 1xsinh13 M1 Makes x the subject. 2 2 4 y must be of the form aebx a ebx c. 1 2 sinh13 ln  3  9 1  ln2 M1 Uses logarithmic form. 4 4 16 y must be of the form aebx a ebx c. 1 2 x2ln2ln4 A1 3 © UCLES 2022 Page 12 of 15

Official mark scheme pages: 12 · source PDF URL

9231-2022-on-21-q05

Oct/Nov 2022 · Paper 21 · Question 5 · 10 marks

5 2m2+2m+1=0m=−11i M1 Correct auxiliary equation. 2 2 y=e−1 2 x(Acos1x+Bsin1x) A1 Complementary function. Accept “y=” 2 2 missing. y= px2 +qx+r leading toy'=2px+q leading to y''=2p B1 Particular integral and its derivatives. p=4 4p+q=3 4p+2q+r=3 M1 Substitutes and equates coefficients. q=−13 r=13 A1 y=e−1 2 x(Acos1x+Bsin1x)+4x2 −13x+13 A1 General solution. 2 2 y'=e−1 2 x(−1 Asin1x+1Bcos1x)−1e−1 2 x(Acos1x+Bsin1x)+8x−13 M1 Full use of product rule. y must be of form 2 2 2 2 2 2 2 shown in the general solution above with p0. A+13=0 1B−1 A−13=0 A=−13,B=13 M1 A1 Uses initial conditions to derive two linear 2 2 equations in two unknowns. y=e−1 2 x(−13cos1x+13sin1x)+4x2−13x+13 A1 2 2 10 © UCLES 2022 Page 9 of 13

Official mark scheme pages: 9 · source PDF URL

9231-2022-on-21-q08

Oct/Nov 2022 · Paper 21 · Question 8 · 14 marks

8(a) du B1 =−2(−1) leading to (−1)d=−1du d 2  −1  1 M1 A1 Applies substitution. For M1, integrand must be   1−(−1)2 d=−1 2   u du=− u +C=− 1−(−1)2 +C of the form k u , where k0 is constant. For A1, allow “+C” missing. 3 8(b) dy y B1 Divides through by . − =sin−1(−1) d  e −−1d =e−ln=−1 M1 A1 Finds integrating factor. d ( −1y ) =sin−1(−1) M1 Correct form on LHS using their integrating factor. d   M1 A1 Integrates sin−1(−1). −1y=sin−1(−1)− d   1−(−1)2    −1  1 M1 Applies part (a) and uses  d=  d+  d 1  1−(−1)2  1−(−1)2  1−(−1)2  1−x2 dx=sin−1x+C. −1y=sin−1(−1)+ 1−(−1)2 −sin−1(−1)+C A1 © UCLES 2022 Page 12 of 13 8(b) 1=1+C M1 Substitutes initial conditions into their expression in y and . y=(−1)sin−1(−1)+ 1−(−1)2 M1 A1 Divides through by their integrating factor. 11 © UCLES 2022 Page 13 of 13

Official mark scheme pages: 12, 13 · source PDF URL

9231-2022-on-22-q04

Oct/Nov 2022 · Paper 22 · Question 4 · 9 marks

4 dx 4 B1 Divides through by4t2 −1 . + x=1 dt 4t2 −1 4 2 2 4 4 1 B1 Writes as partial fractions or a form to which the = − or = = 4t2 −1 2t−1 2t+1 4t2 −1 (2t)2 −1 t2 −(1)2 formula from MF19 can be applied. 2 e 4 ( 4t2−1 )−1 dt =eln(2t−1)−ln(2t+1) = 2t−1 M1 A1 Finds integrating factor. 2t+1 d  2t−1 2t−1 2t+1−2 M1 Correct form on LHS and attempt to integrate RHS. x = = at+b dt 2t+1 2t+1 2t+1 RHS must be of the form with a,b,c,d 0. ct+d 2t−1 A1 x =t−ln(2t+1)+C 2t+1 1=1−ln3+C M1 Substitutes into their expression in x and t. 2t+1  3  M1 A1 After integrating, divides their expression through x=  t+ln   by their coefficient of x. (Their coefficient must 2t−1 2t+1 depend on t.) 9 Question Answer Marks Guidance

Official mark scheme pages: 9 · source PDF URL

9231-2022-on-22-q08

Oct/Nov 2022 · Paper 22 · Question 8 · 14 marks

8(a) dy du B1 =sinhu dx dx d2y d2u du 2 B1 =sinhu +coshu   dx2 dx2 dx d2y dy d2u du 2 du M1 Uses substitution to find y-x equation, AG. + −2y=sinhu +coshu +sinhu −2coshu   dx2 dx dx2 dx dx d2u du du 2 A1 AG cosh2u−1 + +coshu   −2coshu=4e−x dx2 dx  dx © UCLES 2022 Page 13 of 15 8(a) Alternative for 8(a) du dy B1 =(sinhu)−1 dx dx d2u  1 d2y dy coshu du B1 =  −   dx2 sinhudx2 dxsinh2udx d2u  1 d2y dy 2  coshu  = −       dx2 sinhudx2 dx sinh3u sinhu     1   d2y −   dy  2   coshu  +(sinhu)−1 dy + coshu   dy  2 −2coshu M1 U ses substitution to find y-x equation, AG.  sinhudx2 dx sinh3u dx sinh2udx   d2y dy 2  coshu  dy coshu dy 2 − + + −2coshu       dx2 dx sinh2u dx sinh2udx d2y dy A1 AG + −2y=4e−x dx2 dx 4 © UCLES 2022 Page 14 of 15 8(b) m2 +m−2=0m=1,−2 M1 Auxiliary equation. y= Aex +Be−2x A1 Complimentary function. Allow ‘y=’ missing. y=ke−x y'=−ke−x y''=ke−x B1 Particular integral and its derivatives. ke−x −ke−x −2ke−x =4e−x M1 Substitutes and equates coefficients. k=−2 A1 WWW y=coshu= Aex +Be−2x −2e−x A1 Must have ‘y=’ or ‘coshu=’ . du B1 y'=sinhu = Aex −2Be−2x +2e−x dx A+B−2= 5 A−2B+2=4 A= 28, B= 5 M1 A1 Substitutes initial conditions and forms 3 9 9 simultaneous equations. For M1, allow substitution into their y and y' if two linear equations in two unknowns derived. u=cosh−1( 28ex +5e−2x −2e−x) A1 Substitutes for y and finds u in terms of x. 9 9 10 © UCLES 2022 Page 15 of 15

Official mark scheme pages: 13, 14, 15 · source PDF URL

9231-2022-on-23-q05

Oct/Nov 2022 · Paper 23 · Question 5 · 10 marks

5 2m2+2m+1=0m=−11i M1 Correct auxiliary equation. 2 2 y=e−1 2 x(Acos1x+Bsin1x) A1 Complementary function. Accept “y=” 2 2 missing. y= px2 +qx+r leading toy'=2px+q leading to y''=2p B1 Particular integral and its derivatives. p=4 4p+q=3 4p+2q+r=3 M1 Substitutes and equates coefficients. q=−13 r=13 A1 y=e−1 2 x(Acos1x+Bsin1x)+4x2 −13x+13 A1 General solution. 2 2 y'=e−1 2 x(−1 Asin1x+1Bcos1x)−1e−1 2 x(Acos1x+Bsin1x)+8x−13 M1 Full use of product rule. y must be of form 2 2 2 2 2 2 2 shown in the general solution above with p0. A+13=0 1B−1 A−13=0 A=−13,B=13 M1 A1 Uses initial conditions to derive two linear 2 2 equations in two unknowns. y=e−1 2 x(−13cos1x+13sin1x)+4x2−13x+13 A1 2 2 10 © UCLES 2022 Page 9 of 13

Official mark scheme pages: 9 · source PDF URL

9231-2022-on-23-q08

Oct/Nov 2022 · Paper 23 · Question 8 · 14 marks

8(a) du B1 =−2(−1) leading to (−1)d=−1du d 2  −1  1 M1 A1 Applies substitution. For M1, integrand must be   1−(−1)2 d=−1 2   u du=− u +C=− 1−(−1)2 +C of the form k u , where k0 is constant. For A1, allow “+C” missing. 3 8(b) dy y B1 Divides through by . − =sin−1(−1) d  e −−1d =e−ln=−1 M1 A1 Finds integrating factor. d ( −1y ) =sin−1(−1) M1 Correct form on LHS using their integrating factor. d   M1 A1 Integrates sin−1(−1). −1y=sin−1(−1)− d   1−(−1)2    −1  1 M1 Applies part (a) and uses  d=  d+  d 1  1−(−1)2  1−(−1)2  1−(−1)2  1−x2 dx=sin−1x+C. −1y=sin−1(−1)+ 1−(−1)2 −sin−1(−1)+C A1 © UCLES 2022 Page 12 of 13 8(b) 1=1+C M1 Substitutes initial conditions into their expression in y and . y=(−1)sin−1(−1)+ 1−(−1)2 M1 A1 Divides through by their integrating factor. 11 © UCLES 2022 Page 13 of 13

Official mark scheme pages: 12, 13 · source PDF URL

9231-2023-mj-21-q02

May/June 2023 · Paper 21 · Question 2 · 7 marks

2 dy dz B1  1 dx dx dz 13z M1 Substitutes and derives first order separable equation. 1 dx 3z1 dz 13z 6z A1  1 dx 3z1 3z1 1 1z1dz1dx M1 Separates variables and integrates both sides. 2 6 1z1lnzxC1x1 y1ln(xy)C A1 2 6 2 2 6 C1 M1 Substitutes initial conditions into their expression. 2 1ln(x y)1xy1 0 A1 OE. 6 2 2 7 © UCLES 2023 Page 7 of 15

Official mark scheme pages: 7 · source PDF URL

9231-2023-mj-21-q06

May/June 2023 · Paper 21 · Question 6 · 11 marks

6 m2 12m360 M1 Auxiliary equation. xe6tAtB A1 Complementary function. Allow “x” missing. x psintqcostx' pcostqsintx''psintqcost M1 A1 Particular integral and its derivatives. p12q36p37 q12p36q0 M1 Substitutes and equates coefficients. Must have two unknowns. p 35 q12 A1 37 37 xe6tAtB35sint12cost A1 FT Must have “x”. FT on CF. 37 37 x' Ae6t 6e6tAtB35cost12sint M1* Differentiates using product rule. Must have their PI 37 37 differentiated. B12 A6B35 0A1 DM1 A1 Forms simultaneous equations using initial conditions. 37 37 xe6tt12 35sint12cost A1 Must have “x”. 37 37 37 11 © UCLES 2023 Page 12 of 15

Official mark scheme pages: 12 · source PDF URL

9231-2023-mj-22-q02

May/June 2023 · Paper 22 · Question 2 · 7 marks

2 dy dz B1  1 dx dx dz 13z M1 Substitutes and derives first order separable equation. 1 dx 3z1 dz 13z 6z A1  1 dx 3z1 3z1 1 1z1dz1dx M1 Separates variables and integrates both sides. 2 6 1z1lnzxC1x1 y1ln(xy)C A1 2 6 2 2 6 C1 M1 Substitutes initial conditions into their expression. 2 1ln(x y)1xy1 0 A1 OE. 6 2 2 7 © UCLES 2023 Page 7 of 15

Official mark scheme pages: 7 · source PDF URL

9231-2023-mj-22-q06

May/June 2023 · Paper 22 · Question 6 · 11 marks

6 m2 12m360 M1 Auxiliary equation. xe6tAtB A1 Complementary function. Allow “x” missing. x psintqcostx' pcostqsintx''psintqcost M1 A1 Particular integral and its derivatives. p12q36p37 q12p36q0 M1 Substitutes and equates coefficients. Must have two unknowns. p 35 q12 A1 37 37 xe6tAtB35sint12cost A1 FT Must have “x”. FT on CF. 37 37 x' Ae6t 6e6tAtB35cost12sint M1* Differentiates using product rule. Must have their PI 37 37 differentiated. B12 A6B35 0A1 DM1 A1 Forms simultaneous equations using initial conditions. 37 37 xe6tt12 35sint12cost A1 Must have “x”. 37 37 37 11 © UCLES 2023 Page 12 of 15

Official mark scheme pages: 12 · source PDF URL

9231-2023-mj-23-q02

May/June 2023 · Paper 23 · Question 2 · 7 marks

2(a) 6m2 5m10m 1 , 1 M1 Auxiliary equation. 2 3 x Ae1 2 t Be1 3 t A1 Complementary function. Allow with ‘x =’ missing. 2 B1 Particular integral and its derivatives. x pqtrt x'q2rtx''2r 12r5q10rt pqtrt2 t2 10t13 M1 Substitutes and equates coefficients. r 1, q0, p1 A1 x Ae1 2 t Be1 3 t t2 1 A1 Must have x. 6 2(b) xt2 1 B1FT Must have ‘x =’. Accept ‘x is approximately’ but do not accept ‘x‘. FT on their PI. 1 © UCLES 2023 Page 7 of 15

Official mark scheme pages: 7 · source PDF URL

9231-2023-mj-23-q05

May/June 2023 · Paper 23 · Question 5 · 11 marks

5(a) coshx 1  ex ex B1 2 2  ex ex2  1  e2x e2x 2  cosh2x1 M1 A1 Expands, AG. 4 2 3 © UCLES 2023 Page 9 of 15 5(b) e2tanhxdx e2lncoshx cosh2x M1 A1 Finds integrating factor. d  ycosh2x  cosh2x M1 Correct form on LHS, d yI for their integrating dx dx factor I, and attempt to integrate their RHS. ycosh2xcosh2xdx 1 cosh2x1dx 11 sinh2xx  C M1 A1 Uses cosh2 x 1cosh2x1. M1 A1 is for RHS. 2 2 2 2 1C M1 Finds C. ysech2x 1 sinh2x 1 x1  M1 A1 Divides through by their coefficient of y. 4 2 8 © UCLES 2023 Page 10 of 15

Official mark scheme pages: 9, 10 · source PDF URL

9231-2023-on-21-q04

Oct/Nov 2023 · Paper 21 · Question 4 · 10 marks

4 m2 +2m+3=0 M1 Auxiliary equation. [y=]e−x( Acos 2x+Bsin 2x ) A1 Complementary function. Allow with “y=” missing. y= px2 +qx+r y'=2px+q y''=2p B1 Particular integral and its derivatives. 3p=27 3q+4p=0 2p+2q+3r=0 M1 Substitutes and equates coefficients. © UCLES 2023 Page 7 of 14 4 p=9 q=−12 r=2 A1 y=e−x( Acos 2x+Bsin 2x ) +9x2 −12x+2 A1 General solution. Must have “y=”. y'=e−x( − 2Asin 2x+ 2Bcos 2x ) −e−x( Acos 2x+Bsin 2x ) +18x−12 M1* Differentiates. Must use product rule. A+2=2 2B−A−12=−8 DM1 Uses initial conditions A1  A=0,B=2 2 y=2 2e−xsin 2x+9x2 −12x+2 A1 Must have “y=”. 10 Question Answer Marks Guidance

Official mark scheme pages: 7, 8 · source PDF URL

9231-2023-on-21-q06

Oct/Nov 2023 · Paper 21 · Question 6 · 14 marks

6(a) coshx= 1( ex +e−x) sinhx= 1( ex −e−x) B1 2 2 1( ex −e−x)( ex+e−x) = 1( e2x −e−2x) =sinh2x M1 A1 Expands, AG. 2 2 3 © UCLES 2023 Page 9 of 14 6(b) u=sinhxdu=coshxdx B1 sinh22xcoshxdx=4sinh2xcosh2xdu=4sinh2x ( sinh2x+1 ) du M1 Applies identities to find integral in terms of u. =4u2( u2 +1 ) du A1 1 1  1 1  A1 =4  u5 + u3  (+C)=4  sinh5x+ sinh3x  (+C) 5 3  5 3  4 6(c) e tanhxdx =elncoshx =coshx M1 A1 Finds integrating factor. d M1 Correct form on LHS and attempt to integrate RHS. (ycoshx)=sinh22xcoshx dx 1 1  M1 A1 Integrates RHS using their part (b). ycoshx=4  sinh5x+ sinh3x  +C 5 3  4=C M1 Substitutes initial conditions. 1 1  A1 y=4sechx  sinh5x+ sinh3x+1  5 3  7 © UCLES 2023 Page 10 of 14

Official mark scheme pages: 9, 10 · source PDF URL

9231-2023-on-22-q04

Oct/Nov 2023 · Paper 22 · Question 4 · 9 marks

4 e3dx =e3x M1 A1 Finds integrating factor. d ( ye3x) =e3xsinx M1 Correct form on LHS and attempt to integrate RHS. dx EITHER M1 A1 Integrates by parts once or uses e3xsinxdx=−e3xcosx+3e3xcosxdx eix −e-ix sinx= . 2i OR e3xsinxdx=1e3xsinx−1e3xcosxdx 3 3 EITHER M1 Integrates by parts again or e3xsinxdx=−e3xcosx+3 ( e3xsinx−3e3xsinxdx ) eix −e-ix substitutes =sinx and 2i OR ( ) eix +e-ix e3xsinxdx= 1e3xsinx−1 1e3xcosx+1e3xsinxdx =cosx. 3 3 3 3 2 ye3x = 1 e3x(3sinx−cosx)+C A1 Must not see i. 10 1=− 1 +C M1 Finds C. Substitutes into their 10 expression (must be integrated). y= 3sinx− 1 cosx+11e−3x A1 Divides through by coefficient of y. 10 10 10 9 © UCLES 2023 Page 9 of 15

Official mark scheme pages: 9 · source PDF URL

9231-2023-on-22-q08

Oct/Nov 2023 · Paper 22 · Question 8 · 14 marks

8(a) dv dy B1 v= y4 =4y3 dx dx d2v d2y dy 2 B1 =4y3 +12y2   dx2 dx2 dx d2v dv d2y dy 2 dy M1 Uses substitution to find v-x + +4v=4y3 +12y2   +4y3 +4y4 equation, AG. dx2 dx dx2 dx dx  d2y dy 2 dy  A1 AG. =4y3 +3y2   + y3 + y4=4e−2x  dx2 dx dx    Alternative method for question 8(a) dy = 1 dv = 1 dv (B1) y=v 1 4 dx 4y3 dx 4v 3 4 dx d2y 1 d2v 3 dv 2 (B1) = −   dx2 4v 3 4 dx2 16v 7 4 dx v 3 4   1 d2v − 3   dv  2 +3v 1 2   1 dv  2 +v4 3   1 dv +v=e−2x (M1) U eq s u e a s t i s o u n b , s A tit G ut . i on to find v-x  4v 3 4 dx2 16v 7 4 dx   4v 3 4 dx 4v 3 4 dx 1d2v 3 dv 2 3 dv 2 1dv (A1) − + + +v=e−2x     4dx2 16vdx 16vdx 4dx 4 © UCLES 2023 Page 14 of 15 8(b) m2 +m+4=0 M1 Auxiliary equation. v=e−1 2 x( Acos 15 x+Bsin 15 x ) A1 Complimentary function. Allow 2 2 ‘v =’ missing. v=ke−2x v'=−2ke−2x v''=4ke−2x B1 Particular integral and its derivatives. 4ke−2x −2ke−2x +4ke−2x =4e−2x M1 Substitutes and equates coefficients. k = 2 A1 3 v= y4 =e−1 2 x( Acos 15 x+Bsin 15 x ) + 2e−2x A1 Substitutes for y and find v in terms 2 2 3 of x. Must not see i. v'=4y3 dy =e−1 2 x( − 15 Asin 15 x+ 15Bcos 15x ) −1e−1 2 x( Acos 15x+Bsin 15x ) −4e−2x B1 dx 2 2 2 2 2 2 2 3 1= A+ 2 −3 = 15B−1 A−4  A= 1, B=0 M1 A1 Substitutes initial conditions and 3 2 2 2 3 3 forms simultaneous equations. Must have used the product rule when differentiating their v for M1. y= ( 1e−1 2 xcos 15 x+2e−2x )1 4 A1 Accept 3 2 3 y= ( 1e−1 2 xcos 15 x+2e−2x )1 4 3 2 3 10 © UCLES 2023 Page 15 of 15

Official mark scheme pages: 14, 15 · source PDF URL

9231-2023-on-23-q04

Oct/Nov 2023 · Paper 23 · Question 4 · 10 marks

4 m2 +2m+3=0 M1 Auxiliary equation. [y=]e−x( Acos 2x+Bsin 2x ) A1 Complementary function. Allow with “y=” missing. y= px2 +qx+r y'=2px+q y''=2p B1 Particular integral and its derivatives. 3p=27 3q+4p=0 2p+2q+3r=0 M1 Substitutes and equates coefficients. © UCLES 2023 Page 7 of 14 4 p=9 q=−12 r=2 A1 y=e−x( Acos 2x+Bsin 2x ) +9x2 −12x+2 A1 General solution. Must have “y=”. y'=e−x( − 2Asin 2x+ 2Bcos 2x ) −e−x( Acos 2x+Bsin 2x ) +18x−12 M1* Differentiates. Must use product rule. A+2=2 2B−A−12=−8 DM1 Uses initial conditions A1  A=0,B=2 2 y=2 2e−xsin 2x+9x2 −12x+2 A1 Must have “y=”. 10 Question Answer Marks Guidance

Official mark scheme pages: 7, 8 · source PDF URL

9231-2023-on-23-q06

Oct/Nov 2023 · Paper 23 · Question 6 · 14 marks

6(a) coshx= 1( ex +e−x) sinhx= 1( ex −e−x) B1 2 2 1( ex −e−x)( ex+e−x) = 1( e2x −e−2x) =sinh2x M1 A1 Expands, AG. 2 2 3 © UCLES 2023 Page 9 of 14 6(b) u=sinhxdu=coshxdx B1 sinh22xcoshxdx=4sinh2xcosh2xdu=4sinh2x ( sinh2x+1 ) du M1 Applies identities to find integral in terms of u. =4u2( u2 +1 ) du A1 1 1  1 1  A1 =4  u5 + u3  (+C)=4  sinh5x+ sinh3x  (+C) 5 3  5 3  4 6(c) e tanhxdx =elncoshx =coshx M1 A1 Finds integrating factor. d M1 Correct form on LHS and attempt to integrate RHS. (ycoshx)=sinh22xcoshx dx 1 1  M1 A1 Integrates RHS using their part (b). ycoshx=4  sinh5x+ sinh3x  +C 5 3  4=C M1 Substitutes initial conditions. 1 1  A1 y=4sechx  sinh5x+ sinh3x+1  5 3  7 © UCLES 2023 Page 10 of 14

Official mark scheme pages: 9, 10 · source PDF URL

9231-2024-mj-21-q06

May/June 2024 · Paper 21 · Question 6 · 12 marks

6(a) coshxsinhx1 2   1  ex ex  1  ex ex1 2  M1 Substitutes sinh and cosh in terms of 2 2 exponentials.   ex1 2 e 1 2 x A1 AG 2 © Cambridge University Press & Assessment 2024 Page 10 of 14 6(b) m2 m30m1  1i 11 M1 Auxiliary equation. 2 2 ye1 2 x Acos 11xBsin 11x  A1 Complimentary function. 2 2 Allow ‘y’ missing. yke 1 2 x  y' 1ke 1 2 x  y'' 1ke 1 2 x B1 Particular integral and its derivatives. 2 4 1ke 1 2 x 1ke 1 2 x 3ke 1 2 x 5e 1 2 x  1k1k3k 5 M1 Substitutes and equates coefficients. 4 2 4 2 PI must be correct form (aebx). k  4 A1 3 ye1 2 x Acos 11xBsin 11x   4e 1 2 x A1 FT Must have ‘y’. FT on their CF. 2 2 3 dy e1 2 x  11 Asin 11x 11Bcos 11x  1e1 2 x Acos 11xBsin 11x   2e 1 2 x B1 dx 2 2 2 2 2 2 2 3 1 A 4 4  11B1 A 2  A1,B 1 M1 A1 Substitutes initial conditions and forms 3 3 2 2 3 3 11 simultaneous equations. For M1, CF must be correct form ye1 2 x 1cos 11x 1 sin 11x   4e 1 2 x A1 Must have ‘y’. 3 2 11 2 3 10 © Cambridge University Press & Assessment 2024 Page 11 of 14

Official mark scheme pages: 10, 11 · source PDF URL

9231-2024-mj-21-q07

May/June 2024 · Paper 21 · Question 7 · 12 marks

7(a)  x  1 M1 A1 Applies substitution.  dx 1 du u C 1x2 C  1x2 2 u 2 7(b) dy y B1 Divides through by x.  xsinh1x dx x e x1dx elnx x1 M1 A1 Finds integrating factor. d  x1y  sinh1x M1 Correct form on LHS. d Iy for their dx dx integrating factor I.  x M1 A1 Integrates RHS. RHS must be of the form x1y xsinh1x dx  1x2 csinh1x. x1yxsinh1x 1x2 C A1 sinh1xln  x x2 1    *M1 Substitutes initial conditions. 1ln 1 2  2C yx2sinh1xx 1x2 xln  1 2  x  1 2  DM1 A1 Divides through by their integrating factor. Accept yx2sinh1xx 1x2 xsinh11x  1 2  . 10 © Cambridge University Press & Assessment 2024 Page 12 of 14

Official mark scheme pages: 12 · source PDF URL

9231-2024-mj-22-q06

May/June 2024 · Paper 22 · Question 6 · 12 marks

6(a) coshxsinhx1 2   1  ex ex  1  ex ex1 2  M1 Substitutes sinh and cosh in terms of 2 2 exponentials.   ex1 2 e 1 2 x A1 AG 2 © Cambridge University Press & Assessment 2024 Page 10 of 14 6(b) m2 m30m1  1i 11 M1 Auxiliary equation. 2 2 ye1 2 x Acos 11xBsin 11x  A1 Complimentary function. 2 2 Allow ‘y’ missing. yke 1 2 x  y' 1ke 1 2 x  y'' 1ke 1 2 x B1 Particular integral and its derivatives. 2 4 1ke 1 2 x 1ke 1 2 x 3ke 1 2 x 5e 1 2 x  1k1k3k 5 M1 Substitutes and equates coefficients. 4 2 4 2 PI must be correct form (aebx). k  4 A1 3 ye1 2 x Acos 11xBsin 11x   4e 1 2 x A1 FT Must have ‘y’. FT on their CF. 2 2 3 dy e1 2 x  11 Asin 11x 11Bcos 11x  1e1 2 x Acos 11xBsin 11x   2e 1 2 x B1 dx 2 2 2 2 2 2 2 3 1 A 4 4  11B1 A 2  A1,B 1 M1 A1 Substitutes initial conditions and forms 3 3 2 2 3 3 11 simultaneous equations. For M1, CF must be correct form ye1 2 x 1cos 11x 1 sin 11x   4e 1 2 x A1 Must have ‘y’. 3 2 11 2 3 10 © Cambridge University Press & Assessment 2024 Page 11 of 14

Official mark scheme pages: 10, 11 · source PDF URL

9231-2024-mj-22-q07

May/June 2024 · Paper 22 · Question 7 · 12 marks

7(a)  x  1 M1 A1 Applies substitution.  dx 1 du u C 1x2 C  1x2 2 u 2 7(b) dy y B1 Divides through by x.  xsinh1x dx x e x1dx elnx x1 M1 A1 Finds integrating factor. d  x1y  sinh1x M1 Correct form on LHS. d Iy for their dx dx integrating factor I.  x M1 A1 Integrates RHS. RHS must be of the form x1y xsinh1x dx  1x2 csinh1x. x1yxsinh1x 1x2 C A1 sinh1xln  x x2 1    *M1 Substitutes initial conditions. 1ln 1 2  2C yx2sinh1xx 1x2 xln  1 2  x  1 2  DM1 A1 Divides through by their integrating factor. Accept yx2sinh1xx 1x2 xsinh11x  1 2  . 10 © Cambridge University Press & Assessment 2024 Page 12 of 14

Official mark scheme pages: 12 · source PDF URL

9231-2024-mj-23-q05

May/June 2024 · Paper 23 · Question 5 · 10 marks

5(a) m2 10m250m5 M1 Auxiliary equation. xe5tAtB A1 Complementary function. Allow ‘x’ missing. x psintqcostx' pcostqsintx''psintqcost M1 A1 Particular integral and its derivatives. For M1, both derivatives must be present and of the correct form (only allow sign errors). p10q25p338 q10p25q0 M1 Substitutes and equates coefficients. p12 q5 A1 xe5tAtB12sint5cost A1 Must have ‘x’. 7 © Cambridge University Press & Assessment 2024 Page 10 of 15 5(b) x12sint5cost 13 12sint 5 cost  M1 Considering PI only. 13 13 12sint5cost 13sin(t0.395...) A1 A1 A1 for R and A1 for .Condone 22.6. Withhold final A1 mark if working with incorrect PI. CWO. 3 Question Answer Marks Guidance

Official mark scheme pages: 10, 11 · source PDF URL

9231-2024-mj-23-q07

May/June 2024 · Paper 23 · Question 7 · 12 marks

7(a) d x 9 x M1 Applies product rule and dx  2 x2 9 2 cosh1 3   d  cosh1 x  1 dx 3 x2 9 x2 1 9 A1 OE  x2 9 2 x2 9 2 2 x2 9 x2 9 1 A1 Puts over common denominator, AG.  x2 9  x2 9 2 x2 9 2 3 © Cambridge University Press & Assessment 2024 Page 12 of 15 7(b) dy y B1 Divides through by x.   x x2 9 dx x e x1d elnx  x1 M1 A1 Finds integrating factor. d  x1y   x2 9 M1 Correct form on LHS, d Iyfor their dx dx integrating factor I. x 9 x M1 A1 Integrates RHS. For M1, RHS must be of the x1y x2 9 cosh1 C 2 2 3 form c x2 9, where c is a non-zero constant. 1 M1 Substitutes initial conditions. C 3 x 9 x 1 M1 A1 Divides through by their integrating factor. yx  x2 9 cosh1   Accept 2 2 3 3 x 9 x 1  1 yx x2 9 ln   x2 9  . 2 2 3 3  3 9 Question Answer Marks Guidance

Official mark scheme pages: 12, 13 · source PDF URL

9231-2024-on-21-q05

Oct/Nov 2024 · Paper 21 · Question 5 · 10 marks

5 6m2−5m+1=0m=1,1 M1 Auxiliary equation, two distinct real roots. 2 3 x= Ae 1 2 t +Be 1 3 t A1 Complementary function. x= pt2 +qt+rx'=2pt+qx''=2p B1 Particular integral and its derivatives. p=1 −10p+q=1 12p−5q+r=1 M1 Substitutes and equates coefficients. p=1 q=11 r=44 A1 x= Ae 1 2 t +Be 1 3 t +t2 +11t+44 A1 General solution. Need to see ‘x=’. FT on CF. x'=1Ae 1 2 t +1Be 1 3 t +2t+11 M1 Differentiates. 2 3 A+B+44=12 1 A+1B+11=−6 A=−38, B=6 M1A1 Uses initial conditions. 2 3 x=−38e 1 2 t +6e 1 3 t +t2 +11t+44 A1 Need to see ‘x=’. 10 © Cambridge University Press & Assessment 2024 Page 10 of 14

Official mark scheme pages: 10 · source PDF URL

9231-2024-on-21-q07

Oct/Nov 2024 · Paper 21 · Question 7 · 10 marks

7(a) dy + y =x B1 Divides through by x2 +16. dx x2 +16 1 M1A1 Finds integrating factor e  x2+16 dx =e sinh−1( 4 x) A1 AG. = 1x+1 x2 +16 4 4 4 7(b) d ( ( )) M1A1 Correct form on LHS and RHS. y x+ x2 +16 =x2+x x2+16 dx y ( x+ x2 +16 ) =1x3+1 ( x2+16 )3 2 +C M1A1 Integrates RHS. RHS of the correct form. 3 3 6 ( 3+ 25 ) = 27+ 25 25+C M1 Substitutes initial conditions into their expression. 3 3 y ( x+ x2 +16 ) =1x3+1 ( x2 +16 )3 2 −8 A1 OE. 3 3 3 6 © Cambridge University Press & Assessment 2024 Page 12 of 14

Official mark scheme pages: 12 · source PDF URL

9231-2024-on-22-q05

Oct/Nov 2024 · Paper 22 · Question 5 · 10 marks

5 3m2 +2m+1=0m=−1 2i M1 Auxiliary equation. M1 for correct type of 3 3 roots (complex). y=e−1 3 x( Acos 2 x+Bsin 2 x ) A1 Complementary function. 3 3 y= px2 +qx+r y'=2px+q y''=2p B1 Particular integral and its derivatives. p=1 4p+q=0 6p+2q+r=0 M1 Substitutes and equates coefficients. q=−4 r=2 A1 y=e−1 3 x( Acos 2 x+Bsin 2 x ) +x2 −4x+2 A1FT General solution. Must have ‘y=’. FT on 3 3 CF. y'=e−1 3 x( − 2 Asin 2 x+ 2Bcos 2 x ) −1e−1 3 x( Acos 2 x+Bsin 2 x ) +2x−4 *M1 Differentiates. GS must be in correct form. 3 3 3 3 3 3 3 A+2=0 2B−1A−4=0  A=−2,B=5 2 DM1A1 Uses initial conditions. 3 3 y=e−1 3 x( −2cos 2 x+5 2sin 2 x ) +x2 −4x+2 A1 Must have ‘ y=’. 3 3 10 © Cambridge University Press & Assessment 2024 Page 9 of 13

Official mark scheme pages: 9 · source PDF URL

9231-2024-on-22-q07

Oct/Nov 2024 · Paper 22 · Question 7 · 10 marks

7(a) d sech2x M1A1 Applies chain rule. (ln(tanhx))= dx tanhx 1 2 A1 AG. = = . sinhxcoshx sinh2x 3 7(b) dy 2y B1 Divides through by sinh(2x). + =1 dx sinh(2x) d M1A1 Correct form on LHS and RHS. (ytanhx)=tanhx dx ytanhx=ln(coshx)+C M1A1 Integrates RHS. 5 (3) =ln5+C M1 Substitutes initial conditions. 5 4 ytanhx=ln(coshx)+3−ln5 A1 Accept equivalent exact form. 4 7 © Cambridge University Press & Assessment 2024 Page 11 of 13

Official mark scheme pages: 11 · source PDF URL

9231-2024-on-23-q05

Oct/Nov 2024 · Paper 23 · Question 5 · 10 marks

5 6m2 −5m+1=0m=1,1 M1 Auxiliary equation, two distinct real roots. 2 3 x= Ae 1 2 t +Be 1 3 t A1 Complementary function. x= pt2 +qt+rx'=2pt+qx''=2p B1 Particular integral and its derivatives. p=1 −10p+q=1 12p−5q+r=1 M1 Substitutes and equates coefficients. p=1 q=11 r=44 A1 x= Ae 1 2 t +Be 1 3 t +t2 +11t+44 A1 General solution. Need to see ‘x=’. FT on CF. x'=1Ae 1 2 t +1Be 1 3 t +2t+11 M1 Differentiates. 2 3 A+B+44=12 1 A+1B+11=−6 A=−38, B=6 M1A1 Uses initial conditions. 2 3 x=−38e 1 2 t +6e 1 3 t +t2 +11t+44 A1 Need to see ‘x=’. 10 © Cambridge University Press & Assessment 2024 Page 10 of 14

Official mark scheme pages: 10 · source PDF URL

9231-2024-on-23-q07

Oct/Nov 2024 · Paper 23 · Question 7 · 10 marks

7(a) dy y B1 Divides through by x2 +16. + =x dx x2 +16 1 M1A1 Finds integrating factor e  x2+16 dx =e sinh−1( 4 x) A1 AG. = 1x+1 x2 +16 4 4 4 7(b) d ( ( )) M1A1 Correct form on LHS and RHS. y x+ x2 +16 =x2+x x2+16 dx y ( x+ x2 +16 ) =1x3+1 ( x2+16 )3 2 +C M1A1 Integrates RHS. RHS of the correct form. 3 3 ( ) M1 Substitutes initial conditions into their expression. 6 3+ 25 = 27+ 25 25+C 3 3 y ( x+ x2 +16 ) =1x3+1 ( x2 +16 )3 2 −8 A1 OE. 3 3 3 6 © Cambridge University Press & Assessment 2024 Page 12 of 14

Official mark scheme pages: 12 · source PDF URL

9231-2025-mj-21-q05

May/June 2025 · Paper 21 · Question 5 · 10 marks

5 6m2 +3m+6=0m=−1  1i 15 M1 Auxiliary equation. M1 for correct type (complex) of 4 4 roots. x=e−1 4 t( Acos 15t+Bsin 15t ) A1 Complimentary function. 4 4 Allow ‘x=’ missing. Using x instead of t is A0. x=ke−t x'=−ke−t x''=ke−t B1 Particular integral and its derivatives. 6ke−t −3ke−t +6ke−t =e−t M1 Substitutes and equates coefficients. k =1 A1 9 x=e−1 4 t( Acos 15t+Bsin 15t ) +1e−t A1 FT General solution. FT on CF and their value of k. Must 4 4 9 have ‘x=’. x'= 15e−1 4 t( −Asin 15t+Bcos 15t ) −1e−1 4 t( Acos 15t+Bsin 15t ) −1e−t *M1 Using product rule. 4 4 4 4 4 4 9 0= A+1 0= 15 B−1 A−1 A=−1, B= 1 DM1 A1 Substitutes initial conditions and forms simultaneous 9 4 4 9 9 3 15 equations. x=1e−1 4 t( −1cos 15t+ 1 sin 15t ) +1e−t A1 Must have ‘x=’. 3 3 4 15 4 9 10 © Cambridge University Press & Assessment 2025 Page 12 of 18

Official mark scheme pages: 12 · source PDF URL

9231-2025-mj-21-q07

May/June 2025 · Paper 21 · Question 7 · 10 marks

7   x+5 dx= 1ln ( x2 +10x+61 ) M1 A1 Finds integrating factor using  f'(x) dx=ln f(x).  x2 +10x+61 2 f(x) −1ln ( x2+10x+61 ) 1 A1 e 2 = x2 +10x+61 d  y  1 M1 Correct form on LHS using their integrating factor.  =   dx  x2 +10x+61 x2 +10x+61 x2 +10x+61=(x+5)2 +36 B1 Completes the square.  1 x+5 *M1 A1 Integrates RHS. Attempting to integrate by parts is  dx=sinh−1   +C M0.  (x+5)2 +62  6  C=−sinh−1(8) =−ln (8+5) =−ln3 DM1 A1 Substitutes initial conditions. 6 6 3 x2 +1 y 0x+61 =sinh−1    x 6 +5   −sinh−1(4 3 ) A1 Accept x2 +1 y 0x+61 =ln     x 1 + 8 5 + x2 + 1 1 8 0x+61     . 10 © Cambridge University Press & Assessment 2025 Page 16 of 18

Official mark scheme pages: 16 · source PDF URL

9231-2025-mj-22-q05

May/June 2025 · Paper 22 · Question 5 · 10 marks

5 6m2 +3m+6=0m=−1  1i 15 M1 Auxiliary equation. M1 for correct type (complex) of 4 4 roots. x=e−1 4 t( Acos 15t+Bsin 15t ) A1 Complimentary function. 4 4 Allow ‘x=’ missing. Using x instead of t is A0. x=ke−t x'=−ke−t x''=ke−t B1 Particular integral and its derivatives. 6ke−t −3ke−t +6ke−t =e−t M1 Substitutes and equates coefficients. k =1 A1 9 x=e−1 4 t( Acos 15t+Bsin 15t ) +1e−t A1 FT General solution. FT on CF and their value of k. Must 4 4 9 have ‘x=’. x'= 15e−1 4 t( −Asin 15t+Bcos 15t ) −1e−1 4 t( Acos 15t+Bsin 15t ) −1e−t *M1 Using product rule. 4 4 4 4 4 4 9 0= A+1 0= 15 B−1 A−1 A=−1, B= 1 DM1 A1 Substitutes initial conditions and forms simultaneous 9 4 4 9 9 3 15 equations. x=1e−1 4 t( −1cos 15t+ 1 sin 15t ) +1e−t A1 Must have ‘x=’. 3 3 4 15 4 9 10 © Cambridge University Press & Assessment 2025 Page 12 of 18

Official mark scheme pages: 12 · source PDF URL

9231-2025-mj-22-q07

May/June 2025 · Paper 22 · Question 7 · 10 marks

7   x+5 dx= 1ln ( x2 +10x+61 ) M1 A1 Finds integrating factor using  f'(x) dx=ln f(x).  x2 +10x+61 2 f(x) −1ln ( x2+10x+61 ) 1 A1 e 2 = x2 +10x+61 d  y  1 M1 Correct form on LHS using their integrating factor.  =   dx  x2 +10x+61 x2 +10x+61 x2 +10x+61=(x+5)2 +36 B1 Completes the square.  1 x+5 *M1 A1 Integrates RHS. Attempting to integrate by parts is  dx=sinh−1   +C M0.  (x+5)2 +62  6  C=−sinh−1(8) =−ln (8+5) =−ln3 DM1 A1 Substitutes initial conditions. 6 6 3 x2 +1 y 0x+61 =sinh−1    x 6 +5   −sinh−1(4 3 ) A1 Accept x2 +1 y 0x+61 =ln     x 1 + 8 5 + x2 + 1 1 8 0x+61     . 10 © Cambridge University Press & Assessment 2025 Page 16 of 18

Official mark scheme pages: 16 · source PDF URL

9231-2025-mj-23-q04

May/June 2025 · Paper 23 · Question 4 · 10 marks

4 m2 +m−2=0m=1,−2 M1 Auxiliary equation. M1 for correct type of roots. x=Aet +Be−2t A1 Complementary function. Allow “x=” missing. Using x instead of t is A0. x= pt2 +qt+rx'=2pt+qx''=2p B1 Particular integral and its derivatives. p=−1 2p−2q=1 2p+q−2r=−1 M1 Substitutes and equates coefficients. p=−1 q=−3 r=−5 A1 2 4 x= Aet +Be−2t −t2 − 3t− 5 A1 FT General solution. FT on CF and their values 2 4 of p, q and r. Must have “x=”. x'= Aet −2Be−2t −2t− 3 *M1 Differentiates. 2 A+B−5 =0 A−2B−3 =0  A= 4, B=− 1 DM1 A1 Uses initial conditions. 4 2 3 12 x= 4et − 1 e−2t −t2 − 3t− 5 A1 Must have “x=”. 3 12 2 4 10 © Cambridge University Press & Assessment 2025 Page 12 of 17

Official mark scheme pages: 12 · source PDF URL

9231-2025-mj-23-q07

May/June 2025 · Paper 23 · Question 7 · 9 marks

7   2x+6 dx=ln ( x2 +6x+5 ) M1 A1 Finds integrating factor using  x2 +6x+5 f'(x)  dx=ln f(x). f(x) −ln ( x2+6x+5 ) 1 A1 e = x2 +6x+5 d  y  4 M1 Correct form on LHS.  = dxx2 +6x+5 x2 +6x+5 4 1 1 B1 Partial fractions. Or completing the square. = − x2 +6x+5 x+1 x+5  4  x+1 *M1 A1 Integrates RHS. (RHS must be of the correct   x2 +6x+5 dx=ln x+5   +C form.) 0=ln1+C DM1 Substitutes initial conditions. 5 y =ln   x+1 −ln1 =ln   5x+5   A1 Accept y= ( x2 +6x+5 ) ln   5x+5 . x2 +6x+5 x+5 5  x+5   x+5  9 © Cambridge University Press & Assessment 2025 Page 15 of 17

Official mark scheme pages: 15 · source PDF URL

9231-2025-mj-24-q03

May/June 2025 · Paper 24 · Question 3 · 10 marks

3 m2+4m+5=0m=−2i M1 Auxiliary equation. M1 for correct type of roots. y=e−2x(Acosx+Bsinx) A1 Complementary function. Allow ‘y=’ missing. Accept y=e−2x(Acosx+Bisinx). y=ke3x  y'=3ke3x  y''=9ke3x B1 Particular integral and its derivatives. 9k+12k+5k =13k = 1 M1 A1 Substitutes and cancels e3x. 2 y=e−2x(Acosx+Bsinx)+ 1e3x A1 FT General solution. Must have ‘y=’. FT on their CF 2 (The PI must be correct). A+k =1 A=1−k = 1 B1 FT Uses y(0)=1, FT on their 1−k. Their CF must be 2 correct. y'=e−2x(−Asinx+Bcosx)−2e−2x(Acosx+Bsinx)+ 3e3x M1 Finds y'. For M1, their CF must have the correct 2 form. B−2A+ 3 =0B=−1 A1 Uses y'(0)=0. 2 2 y= 1e−2x(cosx−sinx)+ 1e3x A1 Must have ‘y=’. 2 2 10 Question Answer Marks Guidance

Official mark scheme pages: 11 · source PDF URL

9231-2025-mj-24-q06

May/June 2025 · Paper 24 · Question 6 · 9 marks

6 dy 1 B1 Divides through by x. − y=2xtan−1x dx x e −x−1dx =x−1 M1 A1 Finds integrating factor. d *M1 Correct form on LHS and attempt to integrate RHS (yx−1)=2tan−1x dx by parts. Treating tan−1x as cotx is M0. yx−1=2xtan−1x−2   x dx=2xtan−1x−ln ( 1+x2) +C A1 A1 −ln ( 1+x2) =lncos ( tan−1x ) . 1+x2 SC B1 for xtan−1x−1ln ( 1+x2) +C 2 1π=2 (1π ) −ln2+CC=ln2 DM1 Finds C, substitutes given conditions into their 2 4 solution. y=2x2tan−1x−xln ( 1+x2) +xln2 DM1 A1 Divides through by their coefficient of y. 9 © Cambridge University Press & Assessment 2025 Page 14 of 17

Official mark scheme pages: 14 · source PDF URL

9231-2025-on-21-q04

Oct/Nov 2025 · Paper 21 · Question 4 · 10 marks

4 5m2+2m+1=0m=−12i M1 Auxiliary equation. Correct type of roots for M1. 5 5 y=e−1 5 x( Acos2x+Bsin2x ) A1 Complementary function. Allow ‘ y=’ missing. 5 5 y= px2 +qx+r y'=2px+q y''=2p B1 Particular integral and its derivatives. p=1 4p+q=5 10p+2q+r=3 M1 Substitutes and equates coefficients. q=1 r=−9 A1 y=e−1 5 x( Acos2x+Bsin2x ) +x2+x−9 A1 FT General solution. FT on CF. Must have ‘ y=’. 5 5 y'=e−1 5 x( −2Asin2x+2Bcos2x ) −1e−1 5 x( Acos2x+Bsin2x ) +2x+1 *M1 Differentiates using the product rule, CF must be of the 5 5 5 5 5 5 5 correct form. DM1 A1 Uses initial conditions. A−9=0 2B−1 A+1=0 A=9,B=2 5 5 y=e−1 5 x( 9cos2x+2sin2x ) +x2 +x−9 A1 Must have ‘ y=’. 5 5 10 © Cambridge University Press & Assessment 2025 Page 11 of 19

Official mark scheme pages: 11 · source PDF URL

9231-2025-on-21-q07

Oct/Nov 2025 · Paper 21 · Question 7 · 12 marks

7(a) d d x ( tanh−1x ) = 1 2   1 1 + −x x    (1−x)(1 (1 )− − ( x 1 ) + 2 x)(−1) M1 A1 Applies tanh−1x= 1 2 ln    1 1 + −x x   . 1−x 2 1 A1 AG. = 1 =   2 1+x(1−x)2 1−x2 Alternative method for question 7(a) dy M1 A1 Implicit differentiation. tanhy=xsech2 y =1 dx dy 1 1 A1 AG. = = dx 1−tanh2 y 1−x2 3 © Cambridge University Press & Assessment 2025 Page 15 of 19 7(b) dy − y =xtanh−1x B1 Divides through by x. dx x e −x−1d =e−lnx =x−1 M1 A1 Finds integrating factor. d ( x−1y ) =tanh−1x M1 Correct form on LHS and RHS. dx x−1y=xtanh−1x−   x dx *M1 A1 Integrates correct function on RHS. 1−x2 x−1y=xtanh−1x+1ln ( 1−x2) +C A1 ‘+C’ not required. 2 0=1tanh−11+1ln3+CC=−1ln3−1ln3 DM1 Substitutes initial conditions. 2 2 2 4 4 2 4 x−1y=xtanh−1x+1ln ( 1−x2) −1ln27 A1 Accept 2 4 16 x−1y=xtanh−1x+1ln ( 1−x2) −1tanh−11−1ln3. 2 2 2 2 4 Accept y=x ( xtanh−1x+1ln ( 1−x2) −1ln27 ) . 2 4 16 9 © Cambridge University Press & Assessment 2025 Page 16 of 19

Official mark scheme pages: 15, 16 · source PDF URL

9231-2025-on-22-q05

Oct/Nov 2025 · Paper 22 · Question 5 · 10 marks

5(a) m2 +4m+3=0m=−1,−3 M1 Auxiliary equation. This mark could be implied by a correct complementary function. [y =]Ae−x +Be−3x A1 Complementary function. Condone this not being equal to y. x must be used and not another variable. y= psinx+qcosx y'= pcosx−qsinx y''=−psinx−qcosx M1 A1 Particular integral and its derivatives. −p−4q+3p=0 −q+4p+3q=5 M1 Substitutes their particular integral and their derivatives of the correct form and equates coefficients. p =1 q = 1 A1 2 y= Ae−x +Be−3x +sinx+1cosx A1 This must be equal to y. x must be used and not 2 another variable. 7 5(b) y sinx+ 1cosx= 5 ( 2 sinx+ 1 cosx ) M1 e−x →0and for setting up for Pythagoras and 2 2 5 5 applying compound angle formula. Can be implied. 5sin(x+0.464...) A1 A1 A1 for R and A1 for .Condone 26.6. Allow 2 =tan−1 1. OE. 2 Condone decimals for R to at least 3 sf. No method shown award 0 marks. 3 © Cambridge University Press & Assessment 2025 Page 12 of 17

Official mark scheme pages: 12 · source PDF URL

9231-2025-on-22-q07

Oct/Nov 2025 · Paper 22 · Question 7 · 11 marks

7(a) d ( 1 x 4−x2 +2sin−1(1 x) ) =−1 x2( 4−x2)−1 2 + 1 ( 4−x2)1 2 +2 ( 4−x2)−1 2 M1 A1 Applies product rule and formula from MF19 dx 2 2 2 2 for the derivative of sin−1x. = 1 ( 4−x2)( 4−x2)−1 2 + 1 ( 4−x2)1 2 = ( 4−x2)1 2 A1 As a minimum, working of 2 2 1 ( 4−x2)( 4−x2)−1 2 + 1 ( 4−x2)1 2 must be seen 2 2 before the answer. AG. 3 7(b) 1(2+x)−1dx M1 A1 Finds integrating factor. Allow a sign error in e2 = 2+x the integrating factor for the method mark. d ( ) M1 A1 Correct form on LHS and simplifies RHS. y 2+x = 2+x 2−x = 4−x2 dx y 2+x = 1 x 4−x2 +2sin−1(1 x)+C M1 A1 Integrates RHS. 2 2 M1 Substitutes initial conditions. 1 3= 1 3+1π+C 2 2 3 y 2+x = 1 x 4−x2 +2sin−1(1 x)−1π A1 ISW once a correct answer is seen. 2 2 3 8 © Cambridge University Press & Assessment 2025 Page 15 of 17

Official mark scheme pages: 15 · source PDF URL

9231-2025-on-23-q04

Oct/Nov 2025 · Paper 23 · Question 4 · 10 marks

4 5m2+2m+1=0m=−12i M1 Auxiliary equation. Correct type of roots for M1. 5 5 y=e−1 5 x( Acos2x+Bsin2x ) A1 Complementary function. Allow ‘ y=’ missing. 5 5 y= px2 +qx+r y'=2px+q y''=2p B1 Particular integral and its derivatives. p=1 4p+q=5 10p+2q+r=3 M1 Substitutes and equates coefficients. q=1 r=−9 A1 y=e−1 5 x( Acos2x+Bsin2x ) +x2+x−9 A1 FT General solution. FT on CF. Must have ‘ y=’. 5 5 y'=e−1 5 x( −2Asin2x+2Bcos2x ) −1e−1 5 x( Acos2x+Bsin2x ) +2x+1 *M1 Differentiates using the product rule, CF must be of the 5 5 5 5 5 5 5 correct form. DM1 A1 Uses initial conditions. A−9=0 2B−1 A+1=0 A=9,B=2 5 5 y=e−1 5 x( 9cos2x+2sin2x ) +x2 +x−9 A1 Must have ‘ y=’. 5 5 10 © Cambridge University Press & Assessment 2025 Page 11 of 19

Official mark scheme pages: 11 · source PDF URL

9231-2025-on-23-q07

Oct/Nov 2025 · Paper 23 · Question 7 · 12 marks

7(a) d d x ( tanh−1x ) = 1 2   1 1 + −x x    (1−x)(1 (1 )− − ( x 1 ) + 2 x)(−1) M1 A1 Applies tanh−1x= 1 2 ln    1 1 + −x x   . 1−x 2 1 A1 AG. = 1 =   2 1+x(1−x)2 1−x2 Alternative method for question 7(a) dy M1 A1 Implicit differentiation. tanhy=xsech2 y =1 dx dy 1 1 A1 AG. = = dx 1−tanh2 y 1−x2 3 © Cambridge University Press & Assessment 2025 Page 15 of 19 7(b) dy − y =xtanh−1x B1 Divides through by x. dx x e −x−1d =e−lnx =x−1 M1 A1 Finds integrating factor. d ( x−1y ) =tanh−1x M1 Correct form on LHS and RHS. dx x−1y=xtanh−1x−   x dx *M1 A1 Integrates correct function on RHS. 1−x2 x−1y=xtanh−1x+1ln ( 1−x2) +C A1 ‘+C’ not required. 2 0=1tanh−11+1ln3+CC=−1ln3−1ln3 DM1 Substitutes initial conditions. 2 2 2 4 4 2 4 x−1y=xtanh−1x+1ln ( 1−x2) −1ln27 A1 Accept 2 4 16 x−1y=xtanh−1x+1ln ( 1−x2) −1tanh−11−1ln3. 2 2 2 2 4 Accept y=x ( xtanh−1x+1ln ( 1−x2) −1ln27 ) . 2 4 16 9 © Cambridge University Press & Assessment 2025 Page 16 of 19

Official mark scheme pages: 15, 16 · source PDF URL

9231-2025-on-24-q05

Oct/Nov 2025 · Paper 24 · Question 5 · 10 marks

5(a) m2+m+1=0m=−11 3 M1 Auxiliary equation. Must give the correct 2 2 type of roots ab 3. y =e−1 2 x( Acos 3 x+Bsin 3 x ) A1 Complementary function. x missing or 2 2 labelled incorrectly is A0. y= psin2x+qcos2x y'=2pcos2x−2qsin2x y''=−4psin2x−4qcos2x M1 A1 Particular integral and its derivatives. −4p−2q+ p=1 −4q+2p+q=1 M1 Substitutes and equates coefficients. p=− 1 q =− 5 A1 13 13 y =e−1 2 x( Acos 3 x+Bsin 3 x ) − 1 sin2x− 5 cos2x A1 Must have “y =”. 2 2 13 13 7 5(b) y − 1 sin2x− 5 cos2x= 2 ( − 1 sin2x− 5 cos2x ) M1 e−1 2 x →0 and correct use of compound 13 13 13 26 26 angle identity. 2 = 26 . 13 13 2 sin(2x−1.77) A1 A1 A1 for R and A1 for .Condone 101.3. 13 Allow R=0.392 and =cos−1(− 1 ). 26 OE. 3 © Cambridge University Press & Assessment 2025 Page 14 of 19

Official mark scheme pages: 14 · source PDF URL

9231-2025-on-24-q06

Oct/Nov 2025 · Paper 24 · Question 6 · 10 marks

6   1+x dx=−1ln ( 1−2x−x2) M1 A1 Finds integrating factor using 1−2x−x2 2 f '(x)  dx=ln f(x). f(x) −1ln ( 1−2x−x2) 1 A1 e 2 = 1−2x−x2 M1 Correct form on LHS and RHS. d  y  1  = dx  1−2x−x2  1−2x−x2 1−2x−x2 =2−(x+1)2 B1 Completes the square.  1  x+1 *M1 A1 Integrates RHS. RHS must be of the   dx=sin−1   +C 1  2−(x+1)2  2  correct form . ax2 +bx+c π=sin−1 1 +CC = 3π DM1 A1 Substitutes initial conditions. 2 4 y  x+1 A1 =sin−1 + 3π   1−2x−x2  2  4 10 © Cambridge University Press & Assessment 2025 Page 15 of 19

Official mark scheme pages: 15 · source PDF URL

9231-2026-mj-21-q03

May/June 2026 · Paper 21 · Question 3 · 11 marks

3(a) u = c o s h x  d u = s i n h x d x B1  s i n h 2 x d u =  ( u 2 − 1 ) d u M1 A1 Applies correct identity s i n h 2 u = c o s h 2 u − 1 to find integral in terms of u. = 13 u 3 − u ( + C ) =1cosh3x−coshx (+C) A1 3 4 Trace ID: PC-80D328BA4A Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 3(b) © Cambridge University Press & Assessment 2026 Page 11 of 18 e  c o th x d x = e ln sin h x = s i n h x M1 A1 Finds integrating factor. d d x ( y s i n h x ) = s i n h 3 x M1 Correct form on LHS and attempt to integrate RHS. y s i n h x = 13 c o s h 3 x − c o s h x + C *M1 A1 Integrates RHS using their part (a). For M1, their answer to (a) must be of the correct form acosh3x+bcoshx with a,bnon-zero. 0=1(5)3 −5+CC=10 DM1 Substitutes initial conditions. M0 if expression 3 3 3 81 incorrectly manipulated before substitution. E.g. not all terms divided by s i n h x . y = c o s e c h x ( 13 c o s h 3 x − c o s h x + 18 01 ) A1 OE. 7 Question Answer Marks Guidance

Official mark scheme pages: 10, 11 · source PDF URL

9231-2026-mj-21-q06

May/June 2026 · Paper 21 · Question 6 · 10 marks

6 © Cambridge University Press & Assessment 2026 Page 15 of 18 5 m 2 + 4 m + 1 = 0  m = − 25  15 i M1 Auxiliary equation. M1 for correct type of roots abi.  y =  e − 25 x ( A c o s 15 x + B s i n 15 x ) A1 Complementary function. Allow ‘ y = ’ missing. y = p x 2 + q x + r  y ' = 2 p x + q  y '' = 2 p B1 Particular integral and its derivatives. p = 1 8 p + q = 7 10p+4q+r=1 M1 Substitutes and equates coefficients. q = − 1 r = − 5 A1 y = e − 25 x ( A c o s 15 x + B s i n 15 x ) + x 2 − x − 5 A1 General solution. Must have ‘y=’. FT on CF. y ' = e − 25 x ( − 15 A s i n 15 x + 15 B c o s 15 x ) − 25 e − 25 x ( A c o s 15 x + B s i n 15 x ) + 2 x − 1 M1 Differentiates. General solution must be of the correct form. Must use product rule. 0 = A − 5 0= 1B−2 A−1 5 5  A = 5 , B = 1 5 DM1 A1 Uses initial conditions. A1 is both A and B correct. y = e − 25 x ( 5 c o s 15 x + 1 5 s i n 15 x ) + x 2 − x − 5 A1 Must have ‘ y = ’. CWO. 10 Question Answer Marks Guidance

Official mark scheme pages: 15 · source PDF URL

9231-2026-mj-22-q03

May/June 2026 · Paper 22 · Question 3 · 11 marks

3(a) u = c o s h x  d u = s i n h x d x B1  s i n h 2 x d u =  ( u 2 − 1 ) d u M1 A1 Applies correct identity s i n h 2 u = c o s h 2 u − 1 to find integral in terms of u. = 13 u 3 − u ( + C ) =1cosh3x−coshx (+C) A1 3 4 Trace ID: PC-45F835F92B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 3(b) © Cambridge University Press & Assessment 2026 Page 11 of 18 e  c o th x d x = e ln sin h x = s i n h x M1 A1 Finds integrating factor. d d x ( y s i n h x ) = s i n h 3 x M1 Correct form on LHS and attempt to integrate RHS. y s i n h x = 13 c o s h 3 x − c o s h x + C *M1 A1 Integrates RHS using their part (a). For M1, their answer to (a) must be of the correct form acosh3x+bcoshx with a,bnon-zero. 0=1(5)3 −5+CC=10 DM1 Substitutes initial conditions. M0 if expression 3 3 3 81 incorrectly manipulated before substitution. E.g. not all terms divided by s i n h x . y = c o s e c h x ( 13 c o s h 3 x − c o s h x + 18 01 ) A1 OE. 7 Question Answer Marks Guidance

Official mark scheme pages: 10, 11 · source PDF URL

9231-2026-mj-22-q06

May/June 2026 · Paper 22 · Question 6 · 10 marks

6 © Cambridge University Press & Assessment 2026 Page 15 of 18 5 m 2 + 4 m + 1 = 0  m = − 25  15 i M1 Auxiliary equation. M1 for correct type of roots abi.  y =  e − 25 x ( A c o s 15 x + B s i n 15 x ) A1 Complementary function. Allow ‘ y = ’ missing. y = p x 2 + q x + r  y ' = 2 p x + q  y '' = 2 p B1 Particular integral and its derivatives. p = 1 8 p + q = 7 10p+4q+r=1 M1 Substitutes and equates coefficients. q = − 1 r = − 5 A1 y = e − 25 x ( A c o s 15 x + B s i n 15 x ) + x 2 − x − 5 A1 General solution. Must have ‘y=’. FT on CF. y ' = e − 25 x ( − 15 A s i n 15 x + 15 B c o s 15 x ) − 25 e − 25 x ( A c o s 15 x + B s i n 15 x ) + 2 x − 1 M1 Differentiates. General solution must be of the correct form. Must use product rule. 0 = A − 5 0= 1B−2 A−1 5 5  A = 5 , B = 1 5 DM1 A1 Uses initial conditions. A1 is both A and B correct. y = e − 25 x ( 5 c o s 15 x + 1 5 s i n 15 x ) + x 2 − x − 5 A1 Must have ‘ y = ’. CWO. 10 Question Answer Marks Guidance

Official mark scheme pages: 15 · source PDF URL

9231-2026-mj-23-q04

May/June 2026 · Paper 23 · Question 4 · 10 marks

4 © Cambridge University Press & Assessment 2026 Page 11 of 19 4 9 m 2 + 1 4 m + 1 = 0  m = − 17 M1 Auxiliary equation. x = e − 17 t ( A t + B ) A1 Complimentary function. x = k e 17 t  x ' = 17 k e 17 t  x '' = 14 9 k e 17 t B1 Particular integral and its derivatives. k e 17 t + 2 k e 17 t + k e 17 t = 1 9 6 e 17 t  4 k = 1 9 6 M1 Substitutes and equates coefficients. k = 4 9 A1 x = e − 17 t ( A t + B ) + 4 9 e 17 t A1 General solution. x ' = A e − 17 t − 17 e − 17 t ( A t + B ) + 7 e 17 t M1* General solution must be of correct form for any further marks. 7 = B + 4 9 0 = A − 17 B + 7  A = − 1 3 , B = − 4 2 dM1 A1 Substitutes initial conditions to find A and B. x = e − 17 t ( − 1 3 t − 4 2 ) + 4 9 e 17 t A1 10 Trace ID: PC-D4137E1CBF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

Official mark scheme pages: 11 · source PDF URL

9231-2026-mj-23-q06

May/June 2026 · Paper 23 · Question 6 · 10 marks

6(a) © Cambridge University Press & Assessment 2026 Page 14 of 19 d d y x + 1 − y x 2 = 1 − x B1 Divides through by 1 − x 2 . e 1 1 − x 2 d x  = e ta n h −1x o r e 12 ln  11 +− xx   M1 A1 a  dx Finds integrating factor. Allow e1−x2 for M1. 1 1+x A1 AG. Must have obtained all previous marks. ln  1+x =e2 1−x = 1−x 4 6(b) d d x  y 1 1 + − x x  = 1 + x M1 A1 Correct form on LHS and RHS. y 1 1 + − x x = 23 ( 1 + x ) 32 + C M1 A1 Integrates RHS. 1 = 23 + C  C = 13 M1 Substitutes initial conditions. y 1 1 + − x x = 23 ( 1 + x ) 32 + 13 A1 y=2(1+x)(1−x) 1 2 +1(1+x)−1 2(1−x) 1 2 . Final 3 3 answer must be in terms of x and not be in terms of integrating factor. 6 Trace ID: PC-D4137E1CBF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

Official mark scheme pages: 14 · source PDF URL

9231-2026-mj-24-q03

May/June 2026 · Paper 24 · Question 3 · 7 marks

3 EITHER: © Cambridge University Press & Assessment 2026 Page 12 of 19 e  1 d x = e x (M1A1 Finds integrating factor. Allow sign error for M1 e.g. e  − 1 d x . d d x ( y e x ) = e 32 x M1 Correct form on LHS. yex =e 3 2 x = 2e 3 2 x +C *M1A1 Integrates RHS. M1 for integrating RHS of the 3 correct form a e b x . 3 = 23 + C DM1 Substitutes initial conditions. y = 23 e 12 x + 73 e − x A1) OR: Alternative method for 3 y = a e 12 x  y ' = 12 a e 12 x (M1A1 1ae 1 2 x +ae 1 2 x =e 1 2 x a=2 M1A1 2 3 y= 2e 1 2 x +Be−x *M1 3 3 = 23 + B DM1 Substitutes initial conditions. y= 2e 1 2 x +7e−x A1) 3 3 Available marks 7 Trace ID: PC-F3F6C8368E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

Official mark scheme pages: 12 · source PDF URL

9231-2026-mj-24-q08

May/June 2026 · Paper 24 · Question 8 · 13 marks

8(a) © Cambridge University Press & Assessment 2026 Page 18 of 19 d d v x = 3 x y 2 d d y x + y 3 B1 d2v d2y dy dy  dy =3xy2 + 6xy +3y2 +3y2 dx2 dx2 dx dx  dx = 3 x y 2 d d 2 x y 2 + 6 x y  d d y x  2 + 6 y 2 d d y x *B1 Clear use of product and chain rules. 13 d d 2 x v 2 + d d v x + v = x y 2 d d 2 x y 2 + 2 x y  d d y x  2 + 2 y 2 d d y x + 3 x y 2 d d y x + y 3 + x y 3 = x + 5 DM1A1 Substitutes, AG. For M1 must see terms of 13 d d 2 x v 2 separately. 4 Trace ID: PC-F3F6C8368E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 8(b) 1m2+m+1=0 3 © Cambridge University Press & Assessment 2026 Page 19 of 19  m = − 32  32 i  M1 Auxiliary equation. M1 is for the correct type of roots.  v =  e − 32 x ( A c o s 32 x + B s i n 32 x ) A1 Complementary function. Allow ‘ v = ’ missing. [ v = ] a x + b  [ v ' = ] a  [ v '' = ] 0 B1 Particular integral and its derivatives. a + a x + b = x + 5 M1 Substitutes and equates coefficients. a = 1 , b = 4 A1 v = x y 3 = e − 32 x ( A c o s 32 x + B s i n 32 x ) + x + 4 A1 Substitutes for y and find v in terms of x. Must not see i. FT on CF. Must have “ v = ” or ‘ x y 3 = ’. y=3 x−1e−3 2 x( Acos 3x+Bsin 3x ) +1+4x−1 A1 Makes y the subject. 2 2 7 8(c) e − 32 x → 0  y  3 1 + 4 x − 1 M1A1 M1 for stating e−kx →0 (for k0) and deriving 3a+bx−1 . Condone e − 32 x  0 . 2 Trace ID: PC-F3F6C8368E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

Official mark scheme pages: 18, 19 · source PDF URL