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Popular >

y^'+y+1=0

Solution

y^{^{'}} + y + 1 = 0

Solution

y = e^{- t - c_{1}} - 1

Solution steps

y^{'} (t) + y + 1 = 0

Solve separable ODE:

y = e^{- t - c_{1}} - 1

y = e^{- t - c_{1}} - 1

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Frequently Asked Questions (FAQ)

What is the solution for y^'+y+1=0 ?
The solution for y^'+y+1=0 is y=e^{-t-c_{1}}-1