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Popular >

y^'+4y+16=0,y(0)=2

Solution

y^{^{'}} + 4 y + 16 = 0, y (0) = 2

Solution

y = 6 e^{- 4 t} - 4

Solution steps

y^{'} (t) + 4 y + 16 = 0

Solve separable ODE:

y = 6 e^{- 4 t} - 4

y = 6 e^{- 4 t} - 4

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'+4y+16=0,y(0)=2 ?
The solution for y^'+4y+16=0,y(0)=2 is y=6e^{-4t}-4