Question

Solve the initial-value problem y'+2xy=x^2, y(0)=3 and leave the answer in a form involving a definite integral.

Answer 1

@SithsAndGiggles @hartnn

Answer 2

I have the work:

the integrating factor u=e^x^2

(e^x^2)y'+2xy(e^x^2)=(x^2)(e^x^2)

Now what do I do? How do I integrate (x^2)(e^x^2)? I did integration by parts but it only complicated the problem more.

Answer 3

\[\begin{align*}e^{x^2}y'+2xe^{x^2}y&=x^2e^{x^2}\\
\frac{d}{dx}\left[e^{x^2}y\right]&=x^2e^{x^2}\\
e^{x^2}y&=\int x^2e^{x^2}~dx\\
y&=e^{-x^2}\int x^2e^{x^2}~dx\end{align*}\]
The integral doesn't have an elementary result, so you can leave it as is. The final answer will involve a definite integral form, though, to account for the initial value.

Answer 4

Here's some info about how to set up the integral: http://math.mit.edu/suppnotes/suppnotes03/d.pdf

Answer 5

But y(0)=3, when I plug 0 into the y equation, the integral of 0 is 0, right?

Answer 6

For an IVP with initial value \(y(x_0)=y_0\), the solution would have this form:
\[\large e^{x^2}y=y_0+\int_{x_0}^xf(t)~dt\]
so in this case, since \(y(0)=3\), you would use
\[\large e^{x^2}y=3+\int_0^xt^2e^{t^2}~dt=3+\int_0^xt^2e^{t^2}~dt\]
I'm actually not entirely sure about the \(\large e^{x^2}\) where it is... Let's see what happens when you check to see if this solution works.

Differentiate both sides (apply the FTC for the integral):
\[\large e^{x^2}y'+2xe^{x^2}y=0+x^2e^{x^2}\]
which does indeed simplify to the original equation once you divide out the \(e^{x^2}\) factors.

Answer 7

So to summarize, the answer would be
\[\large y=3e^{-x^2}+e^{-x^2}\int_0^xt^2e^{t^2}~dt\]

Answer 8

Thank you so much!

Answer 9

You're welcome!