Question

Solve the initial-value problem xy'+(x+1)y=e^x^2, y(1)=2 and leave the answer in a form involving a definite integral.

Answer 1

Here's the work:

y'+(x+1)/x*y=(e^x^2)/x

the integrating factor u=e^(x+ln(x))=xe^x

(xe^x)y'+e^x(x+1)y=(e^x)(e^x^2)

(xe^x)y=integral of (e^x)(e^x^2)dx

Answer 2

The answer in the book says y=(1/x)(2e^(-(x-1))+e^(-x)*integral of (e^t)(e^t^2)dt from 1 to x)

Answer 3

How do I get the answer from there?

Answer 4

A plot and a possible solution using Mathematica is attached.
Let me be clear, I'm not a diff equation expert.

Answer 5

@SithsAndGiggles

Answer 6

You have the correct general solution,
\[\large xe^xy=\int e^xe^{x^2}~dx\]
With an initial value of \(y(1)=2\), you have
\[\large\color{red}{xe^xy=2+\int_1^x e^{t^2+t}~dt}\]
as the solution.

To check that this works, differentiate the result:
\[\large\begin{align*} \frac{d}{dx}xe^xy&=\frac{d}{dx}\left[2+\int_1^x e^{t^2+t}~dt\right]\\\\
e^xy+xe^xy+xe^x\frac{dy}{dx}&=\frac{d}{dx}\int_1^xe^{t^2+t}~dt\\\\
&=e^{x^2+x}\\\\
xe^x\frac{dy}{dx}&=e^x\left(e^{x^2}-y(x+1)\right)\\\\
\frac{dy}{dx}+\frac{x+1}{x}y&=\frac{e^{x^2}}{x}
\end{align*}\]

Answer 7

I'm pretty sure I linked this page, but in case you lost it: http://math.mit.edu/suppnotes/suppnotes03/d.pdf

Answer 8

So the answer is y=(1/xe^x)(2+integral of e^(t^2+t) dt from 1 to x), right?

Answer 9

I think so. I'm not sure where your book is getting \(\large e^{-(x-1)}\) from though...

Answer 10

That's why it's weird. But the book might be wrong though, because some of the other answers for other problems are wrong.

Answer 11

Thanks a million!

Answer 12

You're welcome!. But now that I think about it, it might be some residual info from a constant of integration, but I'm not sure where it would come into play.