Find functions \({\rm a}(x)\) and \({\rm b}(x)\) such that \(u=x\) and
\(u=\e^{-x}\)
both satisfy the equation
$$\dfrac{\d^2u}{\d x^2} +{\rm a}(x) \dfrac{\d u}{\d x} + {\rm b} (x)u=0\,.$$
For these functions \({\rm a}(x)\) and \({\rm b}(x)\), write
down the general solution of the equation.
Show that the substitution \(y= \dfrac 1 {3u} \dfrac {\d u}{\d x}\)
transforms the equation
\[
\frac{\d y}{\d x} +3y^2 + \frac {x} {1+x} y = \frac 1 {3(1+x)}
\tag{\(*\)}
\]
into
\[
\frac{\d^2 u}{\d x^2} +\frac x{1+x} \frac{\d u}{\d x} - \frac 1 {1+x}
u=0
\]
and hence show that the solution of equation (\(*\)) that satisfies
\(y=0\) at \(x=0\) is given by
\(y = \dfrac{1-\e^{-x}}{3(x+\e^{-x})}\).
Find the solution of the equation
$$
\frac{\d y}{\d x} +y^2 + \frac x {1-x} y = \frac 1 {1-x}
$$
that satisfies \(y=2\) at \(x=0\).
Solve the differential equation
\[
\frac{\mathrm{d}y}{\mathrm{d}x}-y-3y^{2}=-2
\]
by making the substitution \(y=-\dfrac{1}{3u}\dfrac{\mathrm{d}u}{\mathrm{d}x}.\)
Solve the differential equation
\[
x^{2}\frac{\mathrm{d}y}{\mathrm{d}x}+xy+x^{2}y^{2}=1
\]
by making the substitution
\[
y=\frac{1}{x}+\frac{1}{v},
\]
where \(v\) is a function of \(x\).