| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring multiple sophisticated techniques: computing second derivatives with product rule, algebraic manipulation to verify the transformed equation, solving a second-order linear ODE with constant coefficients (complex roots), and back-substituting through the transformation y=x²w. The 11-mark allocation and multi-stage process place it well above average difficulty, though it follows a structured path once the approach is recognized. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks |
|---|---|
| 7(a) | dy dw |
| Answer | Marks | Guidance |
|---|---|---|
| dx dx | B1 | dw 1 dy 2 |
| Answer | Marks | Guidance |
|---|---|---|
| dx2 dx2 dx | B1 | d2w 1 d2y 4 dy 6 |
| Answer | Marks | Guidance |
|---|---|---|
| dx2 dx dx2 dx dx | M1 | Uses substitution to find w-x equation, AG. |
| Answer | Marks | Guidance |
|---|---|---|
| dx2 dx | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| 7(b) | m2 +4m+5=0 leading tom=−2±i | M1 |
| y=e −2x ( Acosx+Bsinx ) | A1 | Complimentary function. Allow “y =” missing. Don’t |
| Answer | Marks | Guidance |
|---|---|---|
| y= px2 +qx+r , y'=2px+q, y''=2p | B1 | Particular integral and its derivatives. |
| Answer | Marks | Guidance |
|---|---|---|
| 5p=5 8p+5q=4 2p+4q+5r=2 | M1 | Substitutes and equates coefficients. |
| Answer | Marks |
|---|---|
| 5 25 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 25 | M1 | Substitutes for y and find w in terms of x. |
| Answer | Marks |
|---|---|
| 4 4 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
--- 7(a) ---
7(a) | dy dw
=x2 +2xw
dx dx | B1 | dw 1 dy 2
= − y
dx x2 dx x3
d2y d2w dw
=x2 +4x +2w
dx2 dx2 dx | B1 | d2w 1 d2y 4 dy 6
= − + y
dx2 x2 dx2 x3 dx x4
d2y dy d2w dw dw
+4 +5y=x2 +4x +2w+4x2 +8xw+5x2w
dx2 dx dx2 dx dx | M1 | Uses substitution to find w-x equation, AG.
d2w ( )dw ( )
5x2 + 4x2 +4x + 5x2 +8x+2 w=5x2 +4x+2
dx2 dx | A1 | AG
4
--- 7(b) ---
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)
5p=5 8p+5q=4 2p+4q+5r=2 | M1 | Substitutes and equates coefficients.
p=1 q=−4 r = 16
5 25 | A1
x2w=e −2x ( Acosx+Bsinx )+x2 −4x+16
5 25 | M1 | Substitutes for y and find w in terms of x.
w= ( xex )−2( Acosx+Bsinx )+1−(5x) −1+ ( 5x )−2
4 4 | A1
7
Question | Answer | Marks | GuidanceIt is given that $y = x^2w$ and
$$x^2\frac{d^2w}{dx^2} + 4x(x + 1)\frac{dw}{dx} + (5x^2 + 8x + 2)w = 5x^2 + 4x + 2.$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 5y = 5x^2 + 4x + 2.$$ [4]
\item Find the general solution for $w$ in terms of $x$. [7]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q7 [11]}}