http://i45.tinypic.com/2lsb9k1.png

I'm not sure. The solutions given are very vague. I think this is the solution to the differential equation.

Thanks.

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- May 21st 2010, 06:08 PMdavismjIs this a solution to the differential equation?
http://i45.tinypic.com/2lsb9k1.png

I'm not sure. The solutions given are very vague. I think this is the solution to the differential equation.

Thanks. - May 21st 2010, 06:19 PMKrizalid
$\displaystyle \frac{y'}{y^3}+\frac1{y^2}=x,$ if $\displaystyle u=\frac1{y^2}\implies u'=-\frac{2y'}{y^2},$ then $\displaystyle -\frac12u'+u=x\implies u'-2u=-2x,$ now $\displaystyle e^{-2x}u'-2e^{-2x}u=-2xe^{-2x},$ and then $\displaystyle \big(u(x)e^{-2x}\big)'=-2xe^{-2x}.$

you can solve the rest now.