Question

Solve y' + 2xy = 2x^3, y(0) = 1.

Answer 1

http://www.sosmath.com/diffeq/first/lineareq/lineareq.html

Answer 2

\[ \huge y = \frac{\int2x^3(e^{\int2x dx})dx +c}{e^{\int2x}}\]

Answer 3

\[y'+2xy=2x^3 \]

Multiply both sides by what is called the integrating factor
It is already in this form:
\[y'+p(x)y=q(x)\]
The integrating factor is :
\[e^{ \int\limits p(x) dx}\]

So for this problem we have the integrating factor is:
\[e^{\int\limits p(x) dx} =e^{\int\limits 2x dx}=e^{x^2}\]

So the reason we multiply by the integrating factor is because it allows us to write the right hand side as
\[(ye^{x^2})'\]
Since this equals :
\[y'e^{x^2} +2xye^{x^2} \text{ By product rule }\]
Which is what we will have after multiplying both sides by the integrating factor
\[e^{x^2}y'+e^{x^2}2xy=e^{x^2} 2x^3\]
So we have again by product rule we can rewrite the right hand "
\[(ye^{x^2})'=e^{x^2} 2x^3\]
Now integrate both sides :)
Don't forget +C after integrating