Show that \(y=\sin^{2}(m\sin^{-1}x)\) satisfies the differential equation
\[
(1-x^{2})y^{(2)}=xy^{(1)}+2m^{2}(1-2y),
\]
and deduce that, for all \(n\geqslant1,\)
\[
(1-x^{2})y^{(n+2)}=(2n+1)xy^{(n+1)}+(n^{2}-4m^{2})y^{(n)},
\]
where \(y^{(n)}\) denotes the \(n\)th derivative of \(y\).
Derive the Maclaurin series for \(y\), making it clear what the general
term is.
By means of the substitution \(x^{\alpha},\) where \(\alpha\) is a suitably chosen constant, find the general solution for \(x>0\) of the differential equation
\[
x\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}-b\frac{\mathrm{d}y}{\mathrm{d}x}+x^{2b+1}y=0,
\]
where \(b\) is a constant and \(b>-1\).
Show that, if \(b>0\), there exist solutions which satisfy \(y\rightarrow1\) and \(\mathrm{d}y/\mathrm{d}x\rightarrow0\) as \(x\rightarrow0\), but that these conditions do not determine a unique solution. For what values of \(b\) do these conditions determine a unique solution?
For \(n=0,1,2,\ldots,\) the functions \(y_{n}\) satisfy the differential equation
\[
\frac{\mathrm{d}^{2}y_{n}}{\mathrm{d}x^{2}}-\omega^{2}x^{2}y_{n}=-(2n+1)\omega y_{n},
\]
where \(\omega\) is a positive constant, and \(y_{n}\rightarrow0\) and \(\mathrm{d}y_{n}/\mathrm{d}x\rightarrow0\) as \(x\rightarrow+\infty\)
and as \(x\rightarrow-\infty.\) Verify that these conditions are satisfied, for \(n=0\) and \(n=1,\) by
\[
y_{0}(x)=\mathrm{e}^{-\lambda x^{2}}\qquad\mbox{ and }\qquad y_{1}(x)=x\mathrm{e}^{-\lambda x^{2}}
\]
for some constant \(\lambda,\) to be determined.
Show that
\[
\frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right)=2(m-n)\omega y_{m}y_{n},
\]
and deduce that, if \(m\neq n,\)
\[
\int_{-\infty}^{\infty}y_{m}(x)y_{n}(x)\,\mathrm{d}x=0.
\]
The functions \(x(t)\) and \(y(t)\) satisfy the simultaneous differential equations
\begin{alignat*}{1}
\dfrac{\mathrm{d}x}{\mathrm{d}t}+2x-5y & =0\\
\frac{\mathrm{d}y}{\mathrm{d}t}+ax-2y & =2\cos t,
\end{alignat*}
subject to \(x=0,\) \(\dfrac{\mathrm{d}y}{\mathrm{d}t}=0\) at \(t=0.\)
Solve these equations for \(x\) and \(y\) in the case when \(a=1\).
Without solving the equations explicitly, state briefly how the form of the solutions for \(x\) and \(y\) if \(a>1\) would differ from the form when \(a=1.\)
Letting \(\mathbf{x} =\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}\) and \(\mathbf{A} = \begin{pmatrix} -2 & 5 \\ -a & 2 \end{pmatrix}\) then our differential equation is \(\mathbf{x}' = \mathbf{Ax} + \begin{pmatrix} 0 \\2 \cos t \end{pmatrix}\).
Looking at the eigenvalues of \(\mathbf{A}\), we find:
\begin{align*}
&& \det \begin{pmatrix} -2-\lambda & 5 \\ -a & 2 -\lambda \end{pmatrix} &= (\lambda^2-4)+5a\\
&&&= \lambda^2 +5a-4
\end{align*}
Therefore if \(a = 1\), \(\lambda = \pm i\).
In which case we should expect the complementary solutions to be of the form \(\mathbf{x} = \begin{pmatrix} A \sin t + B \cos t \\ C \sin t + D \cos t \end{pmatrix}\). The first equation tells us that \((A-5D+B)\cos t + (-B+5C)\sin t=0\) so the complementary solution is:\(\mathbf{x} = \begin{pmatrix} 5(D-C) \sin t + 5C \cos t \\ C \sin t + D \cos t \end{pmatrix}\).
Looking for a particular integral, we should expect to try something like \(\mathbf{x} = \begin{pmatrix} Et\cos t+Ft\sin t\\ Gt\cos t+Ht \sin t\end{pmatrix}\) and we find