seperable equation

**Juggalomike** · December 12th 2010, 01:43 PM

I reviewing for my final and stuck on this question.

dy/dx=y^2/x^2

here is my work

dy/y^2=dx/x^2, integrading both sides i get

-1/y=-1/x+c, i then multiple each side by negative -1 to get
1/y=1/x-c, then multiplying by y i get
1=(1/x-c)y, dividing by 1/x-c i get
1/(1/x-c)=y, but this is not one of the options for my solution, what am i doing wrong?

**dwsmith** · December 12th 2010, 02:19 PM

$\displaystyle \frac{dy}{dx}=\frac{y^2}{x^2}\Rightarrow \int\frac{dy}{y^2}=\int\frac{dx}{x^2}\Rightarrow \frac{-1}{y}=\frac{-1}{x}+C\Rightarrow$

$\displaystyle \frac{-1}{\frac{-1}{y}}=\frac{-1}{\frac{-1+xC}{x}}\Rightarrow y=\frac{-x}{xC-1}$

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