Reduce a 2nd-order ODE using substitution 1

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Further Pure 1 Ex 9B Q2 - the answer is at the bottom of this page

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FP1 Ex 9B Q2 - Generated SBD

$ \newcommand\cosec{{ \,\text{cosec}\, }} \newcommand\fderivy{{ \frac{\text{d}y}{\text{d}x} }} \newcommand\fderivz{{ \frac{\text{d}z}{\text{d}x} }} \newcommand\sderivy{{ \frac{\text{d}^2y}{\text{d}x^2} }} \newcommand\sderivz{{ \frac{\text{d}^2z}{\text{d}x^2} }} $

(a) Show that the transformation $ y = $ ${z \over x} $ transforms the differential equation

\[ x \sderivy + (2 @00 x) \fderivy @02 (@01 @03 x)y = 0 \qquad ...(1) \]

into the differential equation

\[ \sderivz @00 \fderivz @04 z = 0 \qquad ...(2) \]

(6 marks)

(b) Find the general solution to the differential equation (2), giving $z$ as a function of $x$.

(4 marks)

(1 mark)

Solution

\[\begin{align} &\text{(a) First we need to find the first and second derivatives of the substitution given:} \\\\ & z = xy \qquad ...(3) \\\\ & \fderivz = y + x \fderivy \qquad \implies \qquad x \fderivy = \fderivz - y \qquad ...(4) \\\\ & \sderivz = 2 \fderivy + x \sderivy \qquad \implies \qquad x \sderivy = \sderivz - 2 \fderivy \qquad ...(5) \\\\ &\text{Substitute (5) in (1)}. \\\\ & \left( \sderivz - 2 \fderivy \right) + (2 @00 x) \fderivy @02 (@01 @03 x) y = 0 \\\\ &\text{Collect terms:} \\\\ & \sderivz @00 x \fderivy @02 (@01 @03 x) y = 0 \\\\ &\text{Substitute for $x \fderivy$ using (4):} \\\\ & \sderivz @00 \left( \fderivz - y \right) @02 (@01 @03 x) y = 0 \\\\ &\text{Collecting terms again:} \\\\ & \sderivz @00 \fderivz @04 x y = 0 \\\\ &\text{Substitute for $xy$ using (3):} \\\\ & \sderivz @00 \fderivz @04 z = 0 \qquad \text{as required.} \\\\ \\\\ &\text{(b) Write down the complementary function:} \\\\ & t^2 @00 t @04 = 0 \\\\ &\text{We can factorise this:}. \\\\ & (t @05)(t @06) = 0 \\\\ &\text{So the general solution is:} \\\\ & z = A \text{e}^{@07x} + B \text{e}^{@08x} \\\\ \\\\ &\text{(c) Substitute for $z$ using (3):} \\\\ & y = {A \over x} \text{e}^{@07x} + {B \over x} \text{e}^{@08x} \end{align}\]

Answer to the textbook question at the top of this page