Help with general solution to y^2 dy/dx + 2y^2 = 6e^3x

sumrtym77

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Sep 12, 2008
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I studying for a test by working problems and can't seem to get going on this one. Find the general solution to: y^2 dy/dx + 2y^2 = 6e^3x.
So far the farthest I have gotten with this one is : y^2 dy/dx = 6e^3x - 2y^2. Do I need to divide both sides now by y^2. Is this Homogenious First-order?
Please help!
 
Re: Help with general solution

Just to be sure, is this your DE?:

\(\displaystyle y^{2}y'+2y^{2}=6e^{3x}\)
 
Re: Help with general solution

I have only glanced at it, but it looks like we may have a Bernoulli.

If we divide by \(\displaystyle y^{2}\), we get:

\(\displaystyle \frac{dy}{dx}+2=6e^{3x}y^{-2}, \;\ with \;\ n=-2\)

The idea with a Bernoulli is to make the appropriate substitution and turn it into an easier linear DE you can use an integrating factor with.
 
Re: Help with general solution

Well, that makes a difference. That gives \(\displaystyle \frac{dy}{dx}+2y=6e^{3x}y^{-2}\). With n=-2.

Treat it as a Bernoulli and use the sub \(\displaystyle u=y^{3}, \;\ \frac{dy}{dx}=\frac{1}{3}u^{\frac{-2}{3}}\frac{du}{dx}\)

You should be able to transform it into a linear and use the integrating factor.
 
Re: Help with general solution

Thanks galactus, that's waht I done after the hint you gave me when I had the wrong exponent before. I appreciate your help. I wished I would have had this kind of help 3 years ago!
 
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