# Thread: Help finding the general solution

1. ## 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.

2. Well, $\displaystyle n$ doesn't have to positive, does it?

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

So $\displaystyle P(x)=1/3$ and $\displaystyle Q(x)=((e^-^x/3)$?

4. Sure. And what's $\displaystyle n$?

5. -2? But then doesn't a Bernoulli Equation not work? I thought it only works where n=0 or n=1.

6. Your $\displaystyle n$ is correct.

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.

7. 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:
$\displaystyle x/3$?

8. 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:
$\displaystyle rho=x/3$?

9. 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?

10. 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 $\displaystyle w'=0$

11. What does the wiki say that $\displaystyle w$ should be?

12. 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?

13. Right. So what DE does $\displaystyle w$ satisfy?

14. 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.

15. 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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