How to solve this linear ODE with time-varying coefficients? I have a problem with the following ODE of a time-varying system:
$$y''+\frac{4}{t} y' +\frac{2}{t^2} y=u(t), \qquad t>0 $$
The exercises indicates:


*

*Find a representation of the form $x'(t)=A(t)x(t)+B(t)u(t)$

*Find a fundamental solution set and construct a fundamental solutions matrix $X(t)$

*Determine state transition matrix $\phi (t,\tau)$
So, I tried to solve it, my procedure is in these two pictures (text in spanish):
My Procedure part 1
My procedure part 2
I guess it's a total disaster, but I'm confused and I would like any guidance for a better understanding and find the correct procedures. Thanks for your attention.
 A: This is a Euler-Cauchy differential equation. Multiply by $t^2$ to obtain:
$$t^2y’’+4ty’+2y=t^2u.$$
You can determine the solutions of the homogeneous part/fundamental solution by the ansatz $y_h=t^n$. 
$$t^2n(n-1)t^{n-2}+4tnt^{n-1}+2t^{n}=0 \quad \text{; if } t\neq 0\implies n^2+3n+2=0 $$
$$\implies (n+2)(n+1)=0 \implies n_1=-1 \qquad n_2 = -2.$$
After using the ansatz and determining the two possible values for $n_{1,2}$ you can assemble the fundamental system
$$x_1=y=c_1t^{-1}+c_2t^{-2} \implies \boldsymbol{X}_{\text{scalar}}=\begin{bmatrix}t^{-1} & t^{-2} \end{bmatrix}$$
$$x_2=\dot{y}=-c_1t^{-2}-2c_2t^{-3}$$
$$\implies \boldsymbol{x}(t)=c_1\begin{bmatrix}t^{-1} \\-t^{-2} \end{bmatrix}+c_2\begin{bmatrix}t^{-2}\\-2t^{-3} \end{bmatrix}$$
The fundamental solution (columns are vectors associated with the constants of integration $c_1$ and $c_2$) is given by:
$$\boldsymbol{X}(t)=\begin{bmatrix}t^{-1}& t^{-2}\\ t^{-2}&-2t^{-3}\end{bmatrix} \implies \boldsymbol{X}^{-1}(t)= \dfrac{1}{\det\boldsymbol{X}(t)}\begin{bmatrix}-2t^{-3}& -t^{-2}\\ -t^{-2}&t^{-1}\end{bmatrix}.$$

An alternative approach for determining the fundamental system of the matrix equation would be to use the scalar fundamental system $\boldsymbol{X}_{\text{scalar}}$ and the following relationship.
$$\boldsymbol{X}(t)=\begin{bmatrix}\boldsymbol{X}_{\text{scalar}}\\\dot{\boldsymbol{X}}_{\text{scalar}}\end{bmatrix} .$$

The state transition matrix $\boldsymbol{\Phi}(t,\tau)$ is given by:
$$\boldsymbol{\Phi}(t,\tau)=\boldsymbol{X}(t)\boldsymbol{X}^{-1}(\tau).$$

An alternative method to directly obtain the state transition matrix is the Peano-Baker series
$${\boldsymbol {\Phi }}(t,\tau )={\boldsymbol{I}}+\int _{\tau }^{t}{\boldsymbol  {A}}(\sigma _{1})\,d\sigma _{1}+\int _{\tau }^{t}{\boldsymbol  {A}}(\sigma _{1})\int _{\tau }^{{\sigma _{1}}}{\boldsymbol {A}}(\sigma _{2})\,d\sigma _{2}\,d\sigma _{1}$$
$$+\int_{\tau }^{t}{\boldsymbol  {A}}(\sigma _{1})\int _{\tau }^{{\sigma _{1}}}{\boldsymbol  {A}}(\sigma _{2})\int _{\tau }^{{\sigma _{2}}}{\boldsymbol  {A}}(\sigma _{3})\,d\sigma _{3}\,d\sigma _{2}\,d\sigma _{1}+...$$

The general solution is then given by:
$${\boldsymbol  {x}}(t)={\boldsymbol  {\Phi }}(t,t_{0}){\boldsymbol  {x}}(t_{0})+\int _{{t_{0}}}^{t}{\boldsymbol  {\Phi }}(t,\tau ){\boldsymbol  {B}}(\tau ){\boldsymbol  {u}}(\tau )d\tau .$$
As Kwin van der Veen already mentioned your $\boldsymbol{B}(t)$ should be different. In your case it is:
$$\boldsymbol{B}(t)=\begin{bmatrix}0 \\1\end{bmatrix}.$$
A: If you set $t=e^x$ and $z(x)=y(t)=y(e^x)$, then
\begin{align}
z'(x)&=z'(e^x)e^x\\
z''(x)&=y''(e^x)e^{2x}+y'(e^x)e^x\\
&=t^2y''(t)+ty'(t)=t^2zu(t)-3ty'(t)-2y(t)
\end{align}
so that in the end you obtain a linear ODE with constant coefficients
$$
z''(x)+3z'(x)+2z(x)=e^{2x}u(e^x)
$$
which can now be solved with standard methods.
