Find the general solution to the differential equation. Primes denote derivatives with respect tox.

(3y^2)y' + y^3 = e^-x

I beleive this is a Bernoulli Equation but I am still having trouble.

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- Feb 20th 2009, 06:15 AMbearej50DiffyQ
Find the general solution to the differential equation. Primes denote derivatives with respect to

*x*.

(3y^2)y' + y^3 = e^-x

I beleive this is a Bernoulli Equation but I am still having trouble. - Feb 20th 2009, 06:26 AMTheEmptySet

Note that the derivative of $\displaystyle y^3 $ is

$\displaystyle \frac{d}{dx}y^3=3y^2y'$

try the substitution $\displaystyle u=y^3 \implies \frac{du}{dx}=3y^2\frac{dy}{dx}$

So we get the equation $\displaystyle \frac{du}{dx}+u=e^{-x}$

This can be solved with an integrating factor.

Happy Hunting

TES - Feb 20th 2009, 06:46 AMbearej50
I got that far as well. I cannot figure out what the integrating factor would be because there is no

*x*value attached to the*u* - Feb 20th 2009, 06:54 AMTheEmptySet
$\displaystyle \frac{du}{dx}+1\cdot u =e^{-x}$

$\displaystyle p(x)=1 \implies e^{\int 1 dx}=e^x$

$\displaystyle

e^x\frac{du}{dx}+e^xu=1 \iff \frac{d}{dx}\left[ e^x \cdot u\right]=1

$

Now integrating both sides you get

$\displaystyle e^x\cdot u=x+c \implies u=xe^{-x}+ce^{-x} \implies y^3=xe^{-x}+ce^{-x}$