Is this a solution to the differential equation?

Oct 2009
195
19


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

Thanks.
 

Krizalid

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\(\displaystyle \frac{y'}{y^3}+\frac1{y^2}=x,\) 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.
 
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