y' = 2(y^2)x - (e^ -2x)(y^2) , y(0) = 1/2
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Originally Posted by DavidWagner y' = 2(y^2)x - (e^ -2x)(y^2) , y(0) = 1/2 $\displaystyle \frac{dy}{dx} = y^2 \left( 2x - e^{-2x}\right)$ is seperable.
∫(dy/y^2) = ∫2x – e^(-2x)dx -1/y = x^2 + (e^(-2x))/2 + C y(x)= -1/(x^2 + ((e^(-2x)/2))/2 +C y(0) implies C= -2/5 Is this correct?
Originally Posted by DavidWagner ∫(dy/y^2) = ∫2x – e^(-2x)dx -1/y = x^2 + (e^(-2x))/2 + C y(x)= -1/(x^2 + ((e^(-2x)/2))/2 +C y(0) implies C= -2/5 Is this correct? I don't see anything obviously wrong so it's probably OK.
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