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Popular >

y^'+2e^{x+y}=0

Solution

y^{^{'}} + 2 e^{x + y} = 0

Solution

y = - ln (2 e^{x} - c_{1})

Solution steps

y^{'} (x) + 2 e^{x + y} = 0

Solve separable ODE:

y = - ln (2 e^{x} - c_{1})

y = - ln (2 e^{x} - c_{1})

Graph

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Frequently Asked Questions (FAQ)

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