Solve differential equation, problem with integral I have a differential equation: $$(3i*t+2i-3)y''+(2+3t)y'-3y=(3-5i-3i*t)*e^{-t}.$$
First I started looking for a solution to $$(3i*t+2i-3)y''+(2+3t)y'-3y=0.$$
I found a solution: $$y=3*t+2.$$
Then I decided to use the Ostrogradsky–Liouville formula, and got a new equation: $$\left ( \frac{y}{3t+2} \right )' = C* \frac{(3i*t+2i-3)*e^{it}}{(3t+2)^2}.$$
Now I have to solve the integral $$\int \frac{(3i*t+2i-3)*e^{it}}{(3t+2)^2} dt.$$ Here I stopped; I don't know how to do it. WolframAlpha can, but it doesn't show a step-by-step solution, and I must do it myself.
So, can you help me solve the integral, or find another way? Maybe there's a solution without the Ostrogradsky–Liouville formula.
 A: Hint for the integral, try to evaluate the derivative of $\frac {e^{it}}{3t+2}$
A: I'll try to solve only the integral that you're stuck on.
$$\begin{align}I&=\int\dfrac{(3it+2i-3)e^{it}}{(3t+2)^2}\mathrm dt\\&=\int\dfrac{(3t+2)i e^{it}-3e^{it}}{(3t+2)^2}\mathrm dt\\&=\int\dfrac{\mathrm (3t+2)d(e^{it})-\mathrm d(3t+2)\cdot e^{it}}{(3t+2)^2}\mathrm dt\\&=\int\mathrm d\left(\dfrac{e^{it}}{3t+2}\right)\qquad(\text{Recognize the quotient rule?})\\&=\dfrac{e^{it}}{3t+2}+C\end{align}$$
Another approach could be...
$$\begin{align}I&=\int\dfrac{(3it+2i-3)e^{it}}{(3t+2)^2}\mathrm dt\\&=\int\dfrac{ie^{it}}{3t+2}\mathrm dt-\int\dfrac{3e^{it}}{(3t+2)^2}\mathrm dt\end{align}$$We focus on the second part of the integral now, and integrate by parts...
Let $u=-3e^{it}\implies \mathrm du=-3ie^{it}\,\mathrm dt$
And $\mathrm dv=\dfrac1{(3t+2)^2}\mathrm dt\implies v=-\dfrac1{3(3t+2)}$
\begin{align}I&=\int\dfrac{ie^{it}}{3t+2}\mathrm dt+\dfrac{e^{it}}{3t+2}+C-\int\dfrac{ie^{it}}{3t+2}\mathrm dt\\I&=\boxed{\dfrac{e^{it}}{3t+2}+C}\end{align}
