What is the solution for y^{''}+2y^'-15y=4e^{-t} ?

The solution for y^{''}+2y^'-15y=4e^{-t} is y=c_{1}e^{3t}+c_{2}e^{-5t}-1/4 e^{-t}

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y^{''}+2y^'-15y=4e^{-t}

y^{^{''}} + 2 y^{^{'}} - 15 y = 4 e^{- t}

y = c_{1} e^{3 t} + c_{2} e^{- 5 t} - \frac{1}{4} e^{- t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 15 y = 4 e^{- t}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- 5 t} - \frac{1}{4} e^{- t}

y = c_{1} e^{3 t} + c_{2} e^{- 5 t} - \frac{1}{4} e^{- t}

\frac{\partial}{\partial y} (x^{2} - y)

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0

\frac{\partial}{\partial y} (2 x y + z^{2})

\frac{\partial}{\partial x} (3 x ln (y) + y ln (z) - 2 x)

\int x (x + \frac{1}{x}) d x

What is the solution for y^{''}+2y^'-15y=4e^{-t} ?
The solution for y^{''}+2y^'-15y=4e^{-t} is y=c_{1}e^{3t}+c_{2}e^{-5t}-1/4 e^{-t}