# Differential Equation

• September 14th 2008, 03:41 PM
04viking
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)
• September 14th 2008, 04:14 PM
Krizalid
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 $u=\frac1z.$