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Popular >

y^'+2x(y+1)=0

Solution

y^{^{'}} + 2 x (y + 1) = 0

Solution

y = e^{- x^{2} + c_{1}} - 1

Solution steps

y^{'} (x) + 2 x (y + 1) = 0

Solve separable ODE:

y = e^{- x^{2} + c_{1}} - 1

y = e^{- x^{2} + c_{1}} - 1

Graph

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Frequently Asked Questions (FAQ)

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