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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- Sep 14th 2008, 03:41 PM04vikingDifferential 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) - Sep 14th 2008, 04:14 PMKrizalid
2nd. step is wrong.

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