Reduce a 2nd-order ODE using substitution 1
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Generated* SBD with worked solution
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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)
(c) Hence obtain the general solution to differential equation (1).
(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