Year: 2012
Paper: 3
Question Number: 1
Course: UFM Pure
Section: Second order differential equations
The number of candidates attempting more than six questions was, as last year, about 25%, though most of these extra attempts achieved little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Given that $\displaystyle z = y^n \left( \frac{\d y}{\d x}\right)^{\!2}$, show that
\[
\frac{\d z}{\d x} =
y^{n-1} \frac{\d y}{\d x} \left( n \left(\frac{\d y}{\d x}\right)^{\!2} + 2y \frac{\d^2y}{\d x^2}\right)
.
\]
\begin{questionparts}
\item
Use the above result to show that the solution to the equation
\[
\left(\frac{\d y}{\d x}\right)^{\!2}
+ 2y \frac{\d^2y}{\d x^2} = \sqrt y \ \ \ \ \ \ \ \ \ \ (y>0)
\]
that satisfies $y=1$ and $\dfrac{\d y}{\d x}=0$ when $x=0$ is
$y= \big ( \frac38 x^2+1\big)^{\frac23}$.
\item
Find the solution to the equation
\[
\left(\frac{\d y}{\d x}\right)^{\!2}
-y \frac{\d^2y}{\d x^2} + y^2=0
\]
that satisfies $y=1$ and $\dfrac{\d y}{\d x}=0$ when $x=0$.
\end{questionparts}\begin{align*}
&&z &= y^n \left( \frac{\d y}{\d x}\right)^{2} \\
\Rightarrow && \frac{\d z}{\d x} &= ny^{n-1}\left( \frac{\d y}{\d x}\right)^{3} + y^{n} \cdot 2 \left( \frac{\d y}{\d x}\right) \left( \frac{\d^2 y}{\d x^2}\right) \\
&&&= y^{n-1} \left( \frac{\d y}{\d x}\right) \left (n \left( \frac{\d y}{\d x}\right)^2 + 2y \frac{\d^2 y}{\d x^2} \right)
\end{align*}
\begin{questionparts}
\item Let $z = y (y')^2$, then
\begin{align*}
&& \frac{\d z}{\d x} &= y' \sqrt{y} \\
&&&= \sqrt{z} \\
\Rightarrow && \int z^{-1/2} \d z &= x+C \\
\Rightarrow && 2\sqrt{z} &= x + C \\
x = 0, z=0: && C &= 0 \\
\Rightarrow && y(y')^2 &= \frac14 x^2 \\
\Rightarrow &&\sqrt{y} \frac{\d y}{\d x} &= \frac{1}{2}x\\
\Rightarrow && \int \sqrt{y} \d y &= \int \frac{1}{2}x\d x \\
\Rightarrow && \frac23y^{3/2} &=\frac14x^2 + K \\
x = 0, y = 1: && K &= \frac23 \\
\Rightarrow && y &= \left (\frac38 x^2 + 1 \right)^{2/3}
\end{align*}
\item Let $z = y^{-2} (y')^2$
\begin{align*}
&& \frac{\d z}{\d x} &= y^{-3} \frac{\d y}{\d x} \left (-2 \left( \frac{\d y}{\d x}\right) + 2y \frac{\d^2 y}{\d x^2} \right) \\
&&&= y^{-3} \frac{\d y}{\d x} 2y^2 \\
&&&= 2y^{-1}(y') = 2 \sqrt{z} \\
\Rightarrow && 2\sqrt{z} &= 2x + C \\
x = 0, z = 0: && C&= 0 \\
\Rightarrow && z &= x^2 \\
\Rightarrow && \frac{\d y}{\d x} &= xy \\
\Rightarrow && \ln |y| &= \frac12 x^2 + K \\
x =0 , y =1; && K &= 0 \\
\Rightarrow && y &= e^{\frac12 x^2}
\end{align*}
\end{questionparts}
In spite of the printing error at the start of the question, two thirds of the candidates attempted this question. Most candidates earned a quarter of the marks by obtaining z in terms of y in part (i), and then went no further. Some candidates realised the significance of the first line of the question that the expression given was an exact differential, and those that did frequently then scored highly. Some candidates found their own way through having obtained z in terms of y in part (i), then making y the subject substituted back to find a second order differential equation for z, which they then solved and hence completed the solution to each part.