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Popular >

y^'+5xe^y=0

Solution

y^{^{'}} + 5 x e^{y} = 0

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

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n = 0 \sum \infty \frac{6}{n ^{2} - 1}

Frequently Asked Questions (FAQ)

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