What is the solution for y^{''}+4y^'-5y=35e^{-4t} ?

The solution for y^{''}+4y^'-5y=35e^{-4t} is y=c_{1}e^t+c_{2}e^{-5t}-7e^{-4t}

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y^{''}+4y^'-5y=35e^{-4t}

y^{^{''}} + 4 y^{^{'}} - 5 y = 35 e^{- 4 t}

y = c_{1} e^{t} + c_{2} e^{- 5 t} - 7 e^{- 4 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 5 y = 35 e^{- 4 t}

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 5 t} - 7 e^{- 4 t}

y = c_{1} e^{t} + c_{2} e^{- 5 t} - 7 e^{- 4 t}

y^{^{''}} + 3 y = - 4 x^{2} e^{3 x}

y^{^{''}} + 9 y = 27 x^{3}

y^{^{''}} - 4 y = 3 e^{2 x}

\frac{1}{4} y^{^{''}} + 4 y = tan (4 t) - \frac{1}{2} e^{2 t}

y^{^{''''}} + 2 y^{^{''}} + y = 11 + cos (3 t)

What is the solution for y^{''}+4y^'-5y=35e^{-4t} ?
The solution for y^{''}+4y^'-5y=35e^{-4t} is y=c_{1}e^t+c_{2}e^{-5t}-7e^{-4t}