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Popular >

y^'+7xe^y=0

Solution

y^{^{'}} + 7 x e^{y} = 0

Solution

y = - ln (\frac{7 x ^{2}}{2} - c_{1})

Solution steps

y^{'} (x) + 7 x e^{y} = 0

Solve separable ODE:

y = - ln (\frac{7 x ^{2}}{2} - c_{1})

y = - ln (\frac{7 x ^{2}}{2} - c_{1})

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\int_{- 1}^{0} 7 x d x

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\int (e^{x^{2}}) d x

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Frequently Asked Questions (FAQ)

What is the solution for y^'+7xe^y=0 ?
The solution for y^'+7xe^y=0 is y=-ln((7x^2)/2-c_{1})