- #1
snowJT
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Homework Statement
I'm told that this is homogenous
[tex](x^2-xy)y'+y^2 = 0[/tex]
2. The attempt at a solution
This is going to be very painful for me to type out...
[tex](x^2-xy)\frac{dy}{dx} = -y^2[/tex]
[tex]\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}[/tex]
[tex]\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}[/tex]
[tex]v+x\frac{dv}{dx} = \frac{-x^2v^2}{(x^2-x^2v)}[/tex]
[tex]v+x\frac{dv}{dx} = \frac{-x^2v^2}{x^2(1-v)}[/tex]
[tex]v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}[/tex]
[tex]\frac{v+1-v}{-v^2}dv = \frac {dx}{x}[/tex]
[tex]\frac{dv}{-v^2} = \frac {dx}{x}[/tex]
[tex]\int-v^-^2dv = \int\frac {dx}{x}[/tex]
[tex]v^-^1 =ln|x|+C[/tex]
[tex]\frac{y^-^1}{x^-^1} = ln|x|+C[/tex]
[tex]y^-^1 = x ln|x|+C[/tex]
[tex]\frac{1}{y} = x ln|x|+C[/tex]
[tex]\frac{1}{y} = \frac {ln|x|+C}{x}[/tex]
[tex] 0 = \frac {y ln|x|+C}{x}[/tex]
[tex] -Cx = y ln|x|[/tex]
But the answer is...
[tex] xln|y| - y = Cx[/tex]
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