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Popular >

y^'=xe^y,y(0)=0

Solution

y^{^{'}} = x e^{y}, y (0) = 0

Solution

y = - ln (- \frac{x ^{2}}{2} + 1)

Solution steps

y^{'} (x) = x e^{y}

Solve separable ODE:

y = - ln (- \frac{x ^{2}}{2} + 1)

y = - ln (- \frac{x ^{2}}{2} + 1)

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'=xe^y,y(0)=0 ?
The solution for y^'=xe^y,y(0)=0 is y=-ln(-(x^2)/2+1)