CAIE FP1 2010 June — Question 11 EITHER

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeSolve via substitution then back-substitute
DifficultyChallenging +1.3 This is a structured Further Maths question with clear guidance through substitution. While it involves second-order differential equations and requires careful algebraic manipulation of the substitution y=z³, the question explicitly tells students what to show and provides a standard complementary function + particular integral approach. The asymptotic behavior at the end requires some insight but follows naturally from the solution form. Harder than average A-level due to being Further Maths content, but the scaffolding makes it accessible.
Spec4.10e Second order non-homogeneous: complementary + particular integral

The variables \(z\) and \(x\) are related by the differential equation $$3 z ^ { 2 } \frac { \mathrm {~d} ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 6 z ^ { 2 } \frac { \mathrm {~d} z } { \mathrm {~d} x } + 6 z \left( \frac { \mathrm {~d} z } { \mathrm {~d} x } \right) ^ { 2 } + 5 z ^ { 3 } = 5 x + 2$$ Use the substitution \(y = z ^ { 3 }\) to show that \(y\) and \(x\) are related by the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 5 x + 2$$ Given that \(z = 1\) and \(\frac { \mathrm { d } z } { \mathrm {~d} x } = - \frac { 2 } { 3 }\) when \(x = 0\), find \(z\) in terms of \(x\). Deduce that, for large positive values of \(x , z \approx x ^ { \frac { 1 } { 3 } }\).

The variables $z$ and $x$ are related by the differential equation

$$3 z ^ { 2 } \frac { \mathrm {~d} ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 6 z ^ { 2 } \frac { \mathrm {~d} z } { \mathrm {~d} x } + 6 z \left( \frac { \mathrm {~d} z } { \mathrm {~d} x } \right) ^ { 2 } + 5 z ^ { 3 } = 5 x + 2$$

Use the substitution $y = z ^ { 3 }$ to show that $y$ and $x$ are related by the differential equation

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

Given that $z = 1$ and $\frac { \mathrm { d } z } { \mathrm {~d} x } = - \frac { 2 } { 3 }$ when $x = 0$, find $z$ in terms of $x$.

Deduce that, for large positive values of $x , z \approx x ^ { \frac { 1 } { 3 } }$.

\hfill \mbox{\textit{CAIE FP1 2010 Q11 EITHER}}