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As said I'm stuck at the beginning.

$$e^{x^2}y'+2xy=f(x)$$

where $y(0)=2$, $f(x)=\rm I\!R $, and $x = e^{x^2}$ is a solution. Find y. [@Tilper don't edit this part again please]

I've tried everything in my arsenal so far and simply can't find even the homogeneous solution let alone the particular one.

EDIT: $x \mapsto e^{x^2}$ is a solution

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    The homogeneous equation is separable. (And I also don't understand your expression for what $f$ is.) – 2017-02-11
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    $f(x)$ is just a curve in $ \rm I\!R $ any function really. And it lightly confuses me as well. Though Prof. decided it's a must. – 2017-02-11
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    @KatptonLiamfuppinshire Well, as I said, the homogeneous equation is separable... it's just $e^{e^{-x^2}}$ If $f$ is arbitrary then I don't know... you can do a Green function solution or something but it would be expressed as an integral.. What's that $x\to e^{x^2}$ next to $f$ though? it seems to me to be indicating $f(x) = e^{x^2}$ which would be a sensible thing for a homework problem of this sort – 2017-02-11
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    I'm not sure. It only mentions that $x -> e^{x^2}$ is a solution. (Edited the post accordingly) – 2017-02-11
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    @KatptonLiamfuppinshire Well, if it's a solution, then it's the particular solution and you're done. – 2017-02-11

2 Answers 2

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The homogeneous part $$ e^{x^2}y' +2xy = 0$$ can be solved by separation and integration $$ \frac{y'}{y} = -2xe^{-x^2} \\ \ln(y) = e^{-x^2} + C \\y = C'\exp(e^{-x^2}).$$

You are told that $e^{x^2}$ is a particular solution, so the general solution is $$ y(x) = C\exp(e^{-x^2}) + e^{x^2}.$$ Plugging in $y(0)=2$ gives $$ 2= Ce+1$$ so $C=1/e.$

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If you divide by $e^{x^2}$ the DE converts to the form $$y'+P(x)y = e^{-x^2}f(x)$$ where $P(x)=2xe^{-x^2}$. You can then use the formula for an integrating factor $$\mu = e^{\int P(x)dx}$$