Find the solution of the given initial value problem (differential equations) Question: $$\frac{dy}{dt}+ty=1+t$$ where $y(\frac{3}{2})=0$.
So I know this is a non-homogeneous equation and as I started working on it, I ended up with:
$$e^{\frac{t^2}{2}}y=\int_{\frac{3}{2}}^{t} e^{\frac{\tau^2}{2}}d\tau$$
and I'm stuck at this step. Any help?
 A: This is a first order linear differential equation of the form:

The general solution is given by:

where:
 
called the integrating factor.
In your question,
$p(x) =t$ and $q(x) =t+1$.
I guess you can solve it from here.
A: This is a linear first order first degree ODE of the form y'+P(x)y=Q(x).
Here, P(x)=t and Q(x)=1-t.
Now you already found the correct integrating factor(I.F.) i.e. exp(t^2/2).
solution of this form of ode is given by : y*(I.F.)= ∫(I.F.)*Q(x) dx + c , c is a constant of integration
You can easily evaluate this integral without putting the limits of integration and then using the initial value you can find the value of c. 
A: As commented by Frank Lu, this is a very complex differential equation. For $$\frac{dy}{dt}+ty=0$$ the solution is simple $$y=C e^{-\frac{t^2}{2}}$$ and the variation of parameters leads to $$\frac{dC}{dt}=(1+t)e^{\frac{t^2}{2}}$$ so $$C=e^{\frac{t^2}{2}}+\int e^{\frac{t^2}{2}}\,dt+K$$ and the problem comes from the last integral since $$\int e^{\frac{t^2}{2}}\,dt=\sqrt{\frac{\pi }{2}} \text{erfi}\left(\frac{t}{\sqrt{2}}\right)$$ where appears the imaginary error function $\frac{\text{erf(i t)}}{i}$. So, no solution in terms of  elementary functions.
