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For $f: \mathbb{R} \to \mathbb{R}$ differentiable, and inverse function which monotonically increasing. I'd love your help with finding $y$ from the following $f'(y)y'=xf(y)$.

This is what I did so far:

$\frac{f'(y)y'}{f(y)}=x\Rightarrow \frac{f'(y)\frac{dy}{dx}}{f(y)}=x\Rightarrow \int \frac{d(f(y))}{f(y)}=\int\frac{ f'(y)dy}{f(y)}=\int x dx\Rightarrow \ln f(y)=\frac{x^2}{2}$.

What should I do from here?

Thanks a lot!

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So you have solved it. See my coment to the answer on your previus question –  no identity Mar 17 '12 at 13:50

1 Answer 1

up vote 2 down vote accepted

That's pretty obvious, but still: You got $\ln f(y)=\frac{x^2}{2}+C$, hence $f(y)=Ce^{\frac{x^2}{2}}$. Since $f$ has an inverse, $y=f^{-1}\left(Ce^{\frac{x^2}{2}}\right)$

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