Help finding the general solution

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September 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:

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

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:
$(dy/dx) + P(x)y = Q(x)y^n$

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

In it's current state, I can't get it right.
September 16th 2010, 03:24 PM
Ackbeet

Well, $n$ doesn't have to positive, does it?
September 16th 2010, 03:30 PM
Rhode963

$(dy/dx) + y/3 = (1/(3y^2))(e^-^x)$

So $P(x)=1/3$ and $Q(x)=((e^-^x/3)$ ?
September 16th 2010, 03:36 PM
Ackbeet

Sure. And what's $n$ ?
September 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.
September 16th 2010, 03:41 PM
Ackbeet

Your $n$ 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 $n$ is NOT equal to $1$ or $0.$ See here.
September 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:
$y^2+(1/3)y^3=1/(3e^x)$ , so the integrating factor:
$rho=e^3^(^i^n^t^)^1^/^3^d^x$ so does that mean the integrating factor is:
$x/3$ ?
September 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:
$y^2+(1/3)y^3=1/(3e^x)$ , so the integrating factor:
$rho=e^3^\int^1^/^3^d^x$ so does that mean the integrating factor is:
$rho=x/3$ ?
September 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?
September 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 $w$ .

I get $w=(1/3)$ , so that would mean $w'=0$
September 16th 2010, 04:47 PM
Ackbeet

What does the wiki say that $w$ should be?
September 16th 2010, 05:05 PM
Rhode963

Quote:

Originally Posted by Ackbeet

What does the wiki say that $w$ should be?

It says $w$ should be $1/(y^-^3)$ which equals $w=y^3$ ... right?
September 16th 2010, 05:12 PM
Ackbeet

Right. So what DE does $w$ satisfy?
September 16th 2010, 05:17 PM
Rhode963

Quote:

Originally Posted by Ackbeet

Right. So what DE does $w$ satisfy?

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

I know that $w$ is the same as the second term in the original equation... it is the anti-derivative of of the first term.
September 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 $y$ . Instead, the new dependent variable is $w$ . $w$ 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

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

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

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

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