# Help finding the general solution

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• Sep 16th 2010, 03:07 PM
Rhode963
Help finding the general solution
So I've got to find the general solution to the following differential equation:

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

I've been hinted that it is a problem that can be put into a form of a Bernoulli equation, but I can't get it right.

Bernoulli Equations look like this:
$\displaystyle (dy/dx) + P(x)y = Q(x)y^n$

This is what I've tried:
$\displaystyle (3y^2)(dy/dx) + y^3 = e^-^x$
$\displaystyle (dy/dx) + y/3 = (1/(3y^2))(e^-^x)$

In it's current state, I can't get it right.
• Sep 16th 2010, 03:24 PM
Ackbeet
Well, $\displaystyle n$ doesn't have to positive, does it?
• Sep 16th 2010, 03:30 PM
Rhode963
$\displaystyle (dy/dx) + y/3 = (1/(3y^2))(e^-^x)$

So $\displaystyle P(x)=1/3$ and $\displaystyle Q(x)=((e^-^x/3)$?
• Sep 16th 2010, 03:36 PM
Ackbeet
Sure. And what's $\displaystyle n$?
• Sep 16th 2010, 03:38 PM
Rhode963
-2? But then doesn't a Bernoulli Equation not work? I thought it only works where n=0 or n=1.
• Sep 16th 2010, 03:41 PM
Ackbeet
Your $\displaystyle n$ is correct.

Quote:

But then doesn't a Bernoulli Equation not work? I thought it only works where n=0 or n=1.
No. The equation of the form you posted in the OP is only a Bernoulli DE when $\displaystyle n$ is NOT equal to $\displaystyle 1$ or $\displaystyle 0.$ See here.
• Sep 16th 2010, 04:03 PM
Rhode963
Alright, now I don't get the integrating factor. I've gotten my differential equation to the form of:
$\displaystyle y^2+(1/3)y^3=1/(3e^x)$, so the integrating factor:
$\displaystyle rho=e^3^(^i^n^t^)^1^/^3^d^x$ so does that mean the integrating factor is:
$\displaystyle x/3$?
• Sep 16th 2010, 04:05 PM
Rhode963
Alright, now I don't get the integrating factor. I've gotten my differential equation to the form of:
$\displaystyle y^2+(1/3)y^3=1/(3e^x)$, so the integrating factor:
$\displaystyle rho=e^3^\int^1^/^3^d^x$ so does that mean the integrating factor is:
$\displaystyle rho=x/3$?
• Sep 16th 2010, 04:28 PM
Ackbeet
Whoa, slow down, there. I don't think you do the integrating factor yet. You have to do the substitution that renders your DE linear. Go back to the wiki page. What is your substitution, and what is the resulting DE?
• Sep 16th 2010, 04:42 PM
Rhode963
Quote:

Originally Posted by Ackbeet
Whoa, slow down, there. I don't think you do the integrating factor yet. You have to do the substitution that renders your DE linear. Go back to the wiki page. What is your substitution, and what is the resulting DE?

I can't seem to get it to work. I can't get a $\displaystyle w$.

I get $\displaystyle w=(1/3)$, so that would mean $\displaystyle w'=0$
• Sep 16th 2010, 04:47 PM
Ackbeet
What does the wiki say that $\displaystyle w$ should be?
• Sep 16th 2010, 05:05 PM
Rhode963
Quote:

Originally Posted by Ackbeet
What does the wiki say that $\displaystyle w$ should be?

It says $\displaystyle w$ should be $\displaystyle 1/(y^-^3)$ which equals $\displaystyle w=y^3$... right?
• Sep 16th 2010, 05:12 PM
Ackbeet
Right. So what DE does $\displaystyle w$ satisfy?
• Sep 16th 2010, 05:17 PM
Rhode963
Quote:

Originally Posted by Ackbeet
Right. So what DE does $\displaystyle w$ satisfy?

Sorry, I don't understand what you are asking.

I know that $\displaystyle w$ is the same as the second term in the original equation... it is the anti-derivative of of the first term.
• Sep 16th 2010, 05:29 PM
Ackbeet
The whole idea of the Bernoulli equation is that you get to transform your original nonlinear DE into a linear one, which is much easier to solve (you can use integrating factors, for example). The new DE does not have the original dependent variable of $\displaystyle y$. Instead, the new dependent variable is $\displaystyle w$. $\displaystyle w$ is not a solution to the original DE. It is a solution to a related DE, from which you can easily get the solution to the original DE. So for the original DE in the form

$\displaystyle y'+P(x)\,y=Q(x)\,y^{n},$ you substitute $\displaystyle w=1/y^{n-1},$ and the new DE is

$\displaystyle \displaystyle{\frac{w'}{1-n}+P(x)\,w=Q(x).}$

So what's your $\displaystyle P(x)$ and what's your $\displaystyle Q(x)$?
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