Let $$Lu=f$$ be $$a_2(t)\frac{d^2u}{dt^2}+a_1(t)\frac{du}{dt}+a_0(t)u=f$$ Let $u_0$ be a solution for $Lu=0$. Assume that the solution $u$ is of the form $u=u_0v$
a) from the equation $Lu=f$ derive a second order equation for $v$.
My attempt:
$$a_2(t)\frac{d^2u}{dt^2}+a_1(t)\frac{du}{dt}+a_0(t)u=f$$ Now let $u=u_0v$, if we substitute this in the equation we get $$a_2(t)\frac{d^2u_0v}{dt^2}+a_1(t)\frac{du_0v}{dt}+a_0(t)u_0v=f$$ $$a_2(t)[u_0(t)v(t)''+2u_0'(t)v'(t)+u''_0(y)v(t)]+a_1(t)[u_0(t)v'(t)+u'_0(t)v(t)]+a_0(t)u_0(t)v(t)=f$$ $$\frac{d^2v}{dt^2}a_2(t)u_0+\frac{dv}{dt}(2a_2(t)u'_0+a_1(t)u_0)(a_2(t)u''_0+a_1(t)u'_0+a_0(t)u_0) =f$$ $$\frac{d^2v}{dt^2}a_2(t)u_0+\frac{dv}{dt}(2a_2(t)u'_0+a_1(t)u_0)=f$$
So far I think I'm right, but here is where i got stuck
b) let w= v'and solve the differential equation for $w$
My attempt: \If we take $w=\frac{dv}{dt}$, we get $$\frac{dw}{dt}a_2(t)u_0+w(2a_2(t)u'_0+a_1(t)u_0)=f$$ First solve homogenous equation $$\frac{dw}{dt}=\frac{(-2a_2(t)u'_0-a_1(t)u_0)w(t)}{a_2(t)u_0}$$
$$\frac{dw}{dt} \frac{a_2(t)u_0}{w(t)}=2a_2(t)u'_0-a_1(t)u_0$$ View everything except the $w(t)$ as a constant and integrate to $t$ $$\int_{t_0}^t \frac{dw}{dt} \frac{a_2(t)u_0}{w(t)} dt=\int_{t_0}^t 2a_2(t)u'_0-a_1(t)u_0 dt$$ $$a_2(t)u(t)log(w(t))-a_2(t_0)u(t_0)log(w(t_0))=\int_{t_0}^t 2a_2(t)u'_0-a_1(t)u_0dt $$ $$log(w(t))=\frac{\int_{t_0}^t 2a_2(t)u'_0-a_1(t)u_0 dt +a_2(t_0)u(t_0)log(w(t_0))}{a_2(t)u(t)}$$ $$w(t)= \exp\Big[\frac{\int_{t_0}^t 2a_2(t)u'_0-a_1(t)u_0 dt}{a_2(t)u(t)}\Big]\cdot w(t_0)^{\frac{a_2(t_0)u(t_0)}{a_2(t)u(t)}}$$
Which is a horrible equation and its not even with the inhomogenous part
This is a homework question, so I'd like a hint or to be told where I went wrong, nog a full answer
I'm not going to check whether your derivation up to the equation in $w(t)$ is correct.
The equation isn't horrible. It just looks that way when you have so many variables.
Since $a_1(t)$, $a_2(t)$ and $u_0(t)$ are all known functions, let
\begin{align} p(t) &= \frac{2a_2(t)u_0'(t) + a_1(t)u_0(t)}{a_2(t)u_0(t)} \\ q(t) &= \frac{f(t)}{a_2(t)u_0(t)} \end{align}
Then the first order equation reduces to
$$ \frac{dw}{dt} + p(t)w = q(t) $$
You can solve this using the standard integrating factor method, which results in
$$ w(t) = e^{-\int p(t) dt}\int \left[ e^{\int p(t) dt}q(t)\right] dt $$