# Thread: Help finding the general solution

1. Sorry, I know it must be frustrating...
My $P(x)=1/3$ and my $Q(x)=1/(e^x)$

2. Hmm. Your Bernoulli form of the DE looked like this:

$\displaystyle{\frac{dy}{dx} + \frac{1}{3}\,y = \frac{e^{-x}}{3}\,y^{-2},}$ right? I agree with your $P$ but you should double-check your $Q$.

Sorry, I know it must be frustrating...
Believe me, with someone who is putting forth as much effort as you are in order to understand, I'll have all the patience in the world. No need to apologize!

3. Can we back up to $w'$? Is it correct that $w'=3y^2$, but according to the wiki page it is $w'=3y^2y'$

4. According to the wiki, $w$ (not primed) is equal to $y^{3}$ in your case, and therefore $w'=3\,y^{2}\,y'$, by the power and chain rules. This matches up with wiki's

$w'=\dfrac{(1-n)}{y^{n}}\,y',$ as you can see by plugging in $n=-2$. Is that a little clearer now?

5. Yes, so that would give me that same $P(x)$ but $Q(x)=(1/3)e^-^x$?

6. Right. So now your DE for $w$ is:

7. $y^2y'+(y^3)/3=(e^-x)/3$?

8. Oh... I've been plugging in the wrong way... so it's:
$w'/3+w/3=e^-^x/3$

Now what, integrate both sides?

9. Well, probably the first thing to do would be to get rid of all those 3's. Then I'd probably use the integrating factor technique to solve that DE. What do you get when you do that?

10. $w'x+wx=xe^-^x$, then integrate both sides? And if so, with respect to x?

11. Where did all the x's come from? Take your equation in Post # 23 and multiply through by 3. What do you get?

12. $w'+w=e^-^x$. I thought I was putting in the integrating factor... but I think I found the wrong one.

13. Your integrating factor better be an exponential. What do you get?

14. $M(x)=e^3^\int^1^/^3^d^x = e^x$

15. Originally Posted by Rhode963
$M(x)=e^3^\int^1^/^3^d^x = e^x$
Correct. So now multiply everything through by the integrating factor, and write the LHS as the derivative of a product...

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