$2$nd order differential equation - Variation of parameters: $y''-y=\frac{4t^{2}+1}{t\sqrt{t}}$ 
I got a problem solving a $2$nd order differential equation:
$$y''-y=\frac{4t^{2}+1}{t\sqrt{t}}$$

The problem isn't a variation method itself, but this is what I get, when the equations on $c_{1}'(t)$ and $c'_{2}(t)$ are solved:
$$c_{2}'(t)=-2\frac{t^{2}+1}{t\sqrt{t}}e^{t} $$
$$ c'_{1}(t)e^{t}=-c_{2}'(t)e^{-t}$$
How to  integrate the equation for $ c'_{2}(t)$?

My attempt:
The solution for $y''-y=0$ is given by: $$ y=C_{1}e^{t}+C_{2}e^{-t}$$
So by method of variation I get:
$$\left[\begin{array}{cc}e^{t}&e^{-t}\\e^{t}&-e^{-t}\end{array}\right] \left[\begin{array}{c}c_{1}'(t)\\c_{2}'(t)\end{array}\right]=\left[\begin{array}{c}0\\\frac{4t^{2}+1}{t\sqrt{t}}\end{array}\right]$$
Finally I get 2 equations:
$$ \begin{cases} e^{t}c_{1}'(t)+c_{2}'(t)e^{-t}=0\\ e^{t}c_{1}'(t)-c_{2}'(t)e^{-t}=\frac{4t^{2}+1}{t\sqrt{t}}\end{cases} $$
Subtracting these 2 equations, I get:
$$c_{2}'(t)=-2\frac{t^{2}+1}{t\sqrt{t}}e^{t}$$
Which I can't solve.
 A: 
Subtracting these 2 equations, I get:
  $$c_{2}'(t)=-2\frac{t^{2}+1}{t\sqrt{t}}e^{t}$$

That step is incorrect. If you subtract the two equations correctly, you should have:
$$c_2'(t)=-\frac{4t^2+1}{2t\sqrt{t}}e^t$$
Therefore, to evaluate $c_2(t)$, just integrate $c_2'(t)$ with respect to $t$:
$$c_2'(t)=-2t^{1/2}e^t-\frac{1}{2}t^{-3/2}e^t$$
$$c_2(t)=-2\color{red}{\int t^{1/2}e^t~dt}-\frac{1}{2}\color{green}{\int t^{-3/2}e^t~dt} \tag{1}$$

Typically, both integrals on $(1)$ are non-elementary: you could evaluate such integrals using the imaginary error function. However, you can avoid this issue in this case! Let's use integration by parts on the $\color{red}{\text{red}}$ integral:
$$\int t^{1/2}e^t~dt=e^t\cdot t^{1/2}-\frac{1}{2}\int t^{-1/2}e^t~dt$$
Now, on the $\color{green}{\text{green}}$ integral:
$$\int t^{-3/2}e^t~dt=-2t^{-1/2}e^t+2\int t^{-1/2}e^t~dt$$
Combining these two results allows us to cancel the non-elementary integral!
$$\begin{align} c_2(t)&=-2\left(e^t\cdot t^{1/2}-\frac{1}{2}\int t^{-1/2}e^t~dt\right)-\frac{1}{2}\left(-2t^{-1/2}e^t+2\int t^{-1/2}e^t~dt\right)\\&=-2e^{t}t^{1/2}+\int t^{-1/2}e^{t}~dt+t^{-1/2}e^t-\int t^{-1/2}e^{t}~dt\\&=\frac{e^t(1-2t)}{\sqrt{t}}+k_2 \end{align}$$
The same method applies for $c_1(t)$.

If you'd like to verify your answer:
Evaluating $c_1'(t)$ is easy to do using $c_2'(t)$:

 \begin{align} c_1'(t)=\frac{4t^2+1}{2t\sqrt{t}}e^{-t} \end{align}

After integrating $c_1'(t)$ with respect to $t$, you should obtain the following for $c_1(t)$:  

 \begin{align} c_1(t)=-\frac{2t+1}{e^t\sqrt{t}}+k_1 \end{align}  

Therefore, the general solution is:  

 \begin{align} y(t)&=e^t\left(-\frac{2t+1}{e^t\sqrt{t}}+k_1\right)+e^{-t}\left(\frac{e^t(1-2t)}{\sqrt{t}}+k_2\right)\\&=k_1 e^{t}+k_2 e^{-t}-4\sqrt{t} \end{align}

A: Hint:
$$\frac{4t^2+1}{t\sqrt t}e^t=(4t^{1/2}+t^{-3/2})e^t$$ would be fine if it was of the form $(f(t)+f'(t))e^t$.
Now notice that
$$(4t^{1/2}+t^{-3/2})e^t=(4t^{1/2}+2t^{-1/2}-2t^{-1/2}+t^{-3/2})e^t$$ and you have it !
