Question: Use the substitution $x=\frac{1}{t}$ to reduce the differential equation
$$ x^4 \frac{d^2y}{dx^2} + 2x^3 \frac{dy}{dx} + y = t $$
Hence find the general solution of the new equation giving your answer in the form $y=f(t)$
My attempt
$$ x = \frac{1}{t} \Rightarrow x^4 = \frac{1}{t^4} ~~, ~~2x^3 = \frac{2}{t^3}$$
and
$$ x = t^{-1} $$
$$ \Leftrightarrow \frac{dx}{dt} = -t^{-2} $$
$$ \Leftrightarrow dx = \frac{-dt}{t^2} $$
$$ \therefore \frac{1}{dx} = \frac{-t^2}{dt} $$
Again consider
$$\frac{dx}{dt} = -t^{-2} $$
$$ \Leftrightarrow \frac{d^2x}{dt^2} = 2t^{-3} $$
$$ \Leftrightarrow d^2x = \frac{2dt^2}{t^3} $$
$$ \therefore \frac{1}{d^2x} = \frac{t^3}{2dt^2} $$
Now these substitutions transform this differential equation
$$ x^4 \frac{d^2y}{dx^2} + 2x^3 \frac{dy}{dx} + y = t $$
into
$$ \frac{1}{t^4} \cdot d^2y \cdot \frac{t^3}{2dt^2} + \frac{2}{t^3} \cdot dy \cdot\frac{-t^2}{dt} + y = t $$
$$ \frac{1}{2t} \frac{d^2y}{dt^2} - \frac{2}{t} \frac{dy}{dt} + y = t $$
Now I am stuck solving this..
\begin{align*} x &= \frac{1}{t} \\ \dot{x} &= -\frac{1}{t^{2}}\\ y'&= \frac{\dot{y}}{\dot{x}} \\ &= -t^{2} \dot{y} \\ y'' &= \frac{dy'}{dt} \frac{dt}{dx} \\ &= \frac{d}{dt} (-t^{2} \dot{y}) \times (-t^{2}) \\ &= t^{2} (2t\dot{y}+t^{2}\ddot{y}) \\ &= t^{3} (2\dot{y}+t\ddot{y}) \\ t &= \frac{1}{t^{4}} \times t^{3} (2\dot{y}+t\ddot{y})+ \frac{2}{t^{3}} \times (-t^{2} \dot{y})+y \\ t &= \ddot{y}+y \\ y(t) &= t+A\cos t+B\sin t \\ y(x) &= \frac{1}{x}+A\cos \frac{1}{x}+B\sin \frac{1}{x} \end{align*}