What is the solution for y^{''}-2y^'-15y=3e^{2t} ?

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

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y^{''}-2y^'-15y=3e^{2t}

y^{^{''}} - 2 y^{^{'}} - 15 y = 3 e^{2 t}

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 15 y = 3 e^{2 t}

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 5 = 0

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

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

y^{^{''}} - 5 y^{^{'}} + 6 y = cos (x) + 1

y^{^{''}} + 36 y = - 12 sin (6 t)

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