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Math Help - Differential Equation

  1. #1
    Newbie
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    Sep 2008
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    Differential Equation

    I am given the DE y'+y^2 = x^2
    Could anyone verify that i am solving for the general equation correctly?
    dy/dx + y^2 = x^2
    d/dx [(e^x)y^2] = (x^2)e^x
    (e^x)y^2=(x^2)(e^x)-(e^x)+c
    y^2=((x^2)(e^x)-(e^x)+c)/(e^x)
    y={[(x^2)(e^x)-(e^x)+c]/(e^x)}^(1/2)
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    2nd. step is wrong.

    -----

    Put y=xz and the ODE becomes z+xz'+x^{2}z^{2}=x^{2}, which rearranges to z+xz'=x^{2}\left( 1-z^{2} \right)\implies \frac{z'}{z^{2}}+\frac{1}{xz}=x\left( \frac{1}{z^{2}}-1 \right). Finally, put u=\frac1z.
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