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Math Help - Is this a solution to the differential equation?

  1. #1
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    Is this a solution to the differential equation?



    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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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    \frac{y'}{y^3}+\frac1{y^2}=x, if u=\frac1{y^2}\implies u'=-\frac{2y'}{y^2}, then -\frac12u'+u=x\implies u'-2u=-2x, now e^{-2x}u'-2e^{-2x}u=-2xe^{-2x}, and then \big(u(x)e^{-2x}\big)'=-2xe^{-2x}.

    you can solve the rest now.
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