Question: From: Exercise 1.2.5 in Braun's *Differential Equation and their Applications*.

How do we solve $$\frac{dy}{dt} + t^2 y = 1?$$

I cannot evaluate integral
$$
\int e^{\frac{1}{3}t^3} dt.
$$

So I typed this in [Wolfram](http://www.wolframalpha.com/input/?i=int%20e%5E%7B%281/3%29%20t%5E3%7D%20dt) but then... it gets some gamma function..

Answer: 
You can express this integral as a power series: we have
$$
e^{t^3/3}=\sum\_{n=0}^{\infty}\frac{1}{n!}\Bigl(\frac{t^3}{3}\Bigr)^n=\sum\_{n=0}^{\infty}\frac{t^{3n}}{3^n\,n!},\qquad \text{for any }t\in\mathbb{R}.
$$
Then
$$
\int e^{t^3/3}\,dt=C+\sum\_{n=0}^{\infty}\frac{t^{3n+1}}{(3n+1)3^n\,n!},
$$
where $C$ is an arbitrary constant of integration.
Solving a differential equation usually means reducing it to integrals.

Thus, writing the solution as
$$
y=e^{-\frac{t^3}{3}} (\int e^{\frac{t^3}{3}} + C)
$$
is usually good enough.