y' = 2(y^2)x - (e^ -2x)(y^2) , y(0) = 1/2

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- Jan 28th 2009, 12:50 PMDavidWagnerDifferential eqn
y' = 2(y^2)x - (e^ -2x)(y^2) , y(0) = 1/2

- Jan 28th 2009, 01:35 PMmr fantastic
- Jan 28th 2009, 02:05 PMDavidWagner
∫(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? - Jan 28th 2009, 02:13 PMmr fantastic