CAIE FP1 2009 November — Question 9 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2009
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeEuler-Cauchy equations via exponential substitution
DifficultyChallenging +1.2 This is a structured Euler-Cauchy equation problem with guided steps: proving a given substitution formula, applying it to transform the equation, then solving a constant-coefficient DE. While it requires careful chain rule manipulation and knowledge of auxiliary equations with particular integrals, the question provides the key substitution and intermediate result, making it more accessible than discovering the method independently. It's harder than routine single-method questions but easier than unguided proof or multi-insight problems.
Spec4.10c Integrating factor: first order equations4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral
9 Show that if \(y\) depends on \(x\) and \(x = \mathrm { e } ^ { u }\) then $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} u } .$$ Given that \(y\) satisfies the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = 30 x ^ { 2 }$$ use the substitution \(x = \mathrm { e } ^ { u }\) to show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} u } + 3 y = 30 \mathrm { e } ^ { 2 u }$$ Hence find the general solution for \(y\) in terms of \(x\).
AnswerMarks
\(x\frac{dy}{dx} = \frac{dy}{du}\)B1
Some relevant, correct intermediate result such as: \(x^2(d^2y/dx^2) + x(dy/dx) = d^2y/du^2\)M1A1
\(\Rightarrow x^2\frac{d^2y}{dx} = \frac{d^2y}{du^2} - \frac{dy}{du}\) (AG)A1
Uses \(x = e^u\) to obtain:
\(\frac{d^2y}{du^2} + 4\frac{dy}{du} + 3y = 30e^{2u}\) (AG)M1A1
Complementary function \(= Ae^{-u} + Be^{-3u}\)M1A1
Particular integral \(= 2e^{2u}\)M1A1
General solution for \(y\) in the \(x\)-domain: \(y = A/x + B/x^3 + 2x^2\)A1 
$x\frac{dy}{dx} = \frac{dy}{du}$ | B1 |
Some relevant, correct intermediate result such as: $x^2(d^2y/dx^2) + x(dy/dx) = d^2y/du^2$ | M1A1 |
$\Rightarrow x^2\frac{d^2y}{dx} = \frac{d^2y}{du^2} - \frac{dy}{du}$ (AG) | A1 |
Uses $x = e^u$ to obtain: | |
$\frac{d^2y}{du^2} + 4\frac{dy}{du} + 3y = 30e^{2u}$ (AG) | M1A1 |
Complementary function $= Ae^{-u} + Be^{-3u}$ | M1A1 |
Particular integral $= 2e^{2u}$ | M1A1 |
General solution for $y$ in the $x$-domain: $y = A/x + B/x^3 + 2x^2$ | A1 |
9 Show that if $y$ depends on $x$ and $x = \mathrm { e } ^ { u }$ then

$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} u } .$$

Given that $y$ satisfies the differential equation

$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = 30 x ^ { 2 }$$

use the substitution $x = \mathrm { e } ^ { u }$ to show that

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} u ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} u } + 3 y = 30 \mathrm { e } ^ { 2 u }$$

Hence find the general solution for $y$ in terms of $x$.

\hfill \mbox{\textit{CAIE FP1 2009 Q9 [11]}}
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