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Popular >

y^'+4e^{x+y}=0

Solution

y^{^{'}} + 4 e^{x + y} = 0

Solution

y = - ln (4 e^{x} - c_{1})

Solution steps

y^{'} (x) + 4 e^{x + y} = 0

Solve separable ODE:

y = - ln (4 e^{x} - c_{1})

y = - ln (4 e^{x} - c_{1})

Popular Examples

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\int \frac{1}{t ^{2} 441 - t ^{2}} d t

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\int (4 x^{5} + 1) x^{4} d x

(\partial)/(\partial x)(cos^3(x^7y^3))

\frac{\partial}{\partial x} (cos^{3} (x^{7} y^{3}))

Frequently Asked Questions (FAQ)

What is the solution for y^'+4e^{x+y}=0 ?
The solution for y^'+4e^{x+y}=0 is y=-ln(4e^x-c_{1})