find the general solution of the first order diff. equation. e^y + 2xy + (xe^y + x^2)dy/dy = 0 I divided accross by (xe^y + x^2) to get 2ln(y) - ln(x) + dy/dx = 0 Not sure where to take it from here.Help would be greatly appreciated.
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Originally Posted by matty888 find the general solution of the first order diff. equation. e^y + 2xy + (xe^y + x^2)dy/dy = 0 I divided accross by (xe^y + x^2) to get 2ln(y) - ln(x) + dy/dx = 0 Not sure where to take it from here.Help would be greatly appreciated. I think what you meant is Really not sure what you did but multiplying by dx gives and this ODE is exact meaning that there exist a such that
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