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Popular >

y^{''}+4y^'+8=0

Solution

y^{^{''}} + 4 y^{^{'}} + 8 = 0

Solution

y = c_{1} + c_{2} e^{- 4 t} - 2 t

Solution steps

y^{''} (t) + 4 y^{'} (t) + 8 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} - 2 t

y = c_{1} + c_{2} e^{- 4 t} - 2 t

Popular Examples

y^{''}-4y^'-12y=10e^{-2x}

y^{^{''}} - 4 y^{^{'}} - 12 y = 10 e^{- 2 x}

y^{^{''}} + 8 y = 8 x

y^{'''}+12y^{''}+48y^'+64y=8e^{-4x}

y^{^{'''}} + 12 y^{^{''}} + 48 y^{^{'}} + 64 y = 8 e^{- 4 x}

y^{''}+y=8e^{-6x},y(0)=0,y^'(0)=0

y^{^{''}} + y = 8 e^{- 6 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{''}-6y^'-6y=2xe^{-x}

y^{^{''}} - 6 y^{^{'}} - 6 y = 2 x e^{- x}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+4y^'+8=0 ?
The solution for y^{''}+4y^'+8=0 is y=c_{1}+c_{2}e^{-4t}-2t