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

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

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

y^{^{'''}} + 3 y^{^{''}} + 2 y^{^{'}} = u

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x^{2} - 2 x + 1 + sin (x)

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 x + 6

y^{^{''}} - 2 y^{^{'}} - 8 y = 3 e^{3 x}

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