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Popular >

y^'=5xe^y,y(0)=0

Solution

y^{^{'}} = 5 x e^{y}, y (0) = 0

Solution

y = - ln (- \frac{5 x ^{2}}{2} + 1)

Solution steps

y^{'} (x) = 5 x e^{y}

Solve separable ODE:

y = - ln (- \frac{5 x ^{2}}{2} + 1)

y = - ln (- \frac{5 x ^{2}}{2} + 1)

Graph

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\int_{- 2}^{2} \frac{1}{( x + 4 ) ( x - 4 )} d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=5xe^y,y(0)=0 ?
The solution for y^'=5xe^y,y(0)=0 is y=-ln(-(5x^2)/2+1)