# DiffyQ

• Feb 20th 2009, 06:15 AM
bearej50
DiffyQ
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 AM
TheEmptySet
Quote:

Originally Posted by bearej50
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.

Note that the derivative of $y^3$ is
$\frac{d}{dx}y^3=3y^2y'$

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

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

This can be solved with an integrating factor.

Happy Hunting

TES
• Feb 20th 2009, 06:46 AM
bearej50
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 AM
TheEmptySet
$\frac{du}{dx}+1\cdot u =e^{-x}$

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

$
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

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