What is the solution for y^{''}-4y^'+3y=96e^{-5t}-24e^{-3t}+18 ?

The solution for y^{''}-4y^'+3y=96e^{-5t}-24e^{-3t}+18 is y=c_{1}e^{3t}+c_{2}e^t+2e^{-5t}-e^{-3t}+6

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y^{''}-4y^'+3y=96e^{-5t}-24e^{-3t}+18

y^{^{''}} - 4 y^{^{'}} + 3 y = 96 e^{- 5 t} - 24 e^{- 3 t} + 18

y = c_{1} e^{3 t} + c_{2} e^{t} + 2 e^{- 5 t} - e^{- 3 t} + 6

Solution steps

y^{''} (t) - 4 y^{'} (t) + 3 y = 96 e^{- 5 t} - 24 e^{- 3 t} + 18

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{t} + 2 e^{- 5 t} - e^{- 3 t} + 6

y = c_{1} e^{3 t} + c_{2} e^{t} + 2 e^{- 5 t} - e^{- 3 t} + 6

y^{^{''}} + 6 y^{^{'}} + 9 y = x e^{2 x}

x^{^{''}} + 4 x^{^{'}} + 6 x = 1 + e^{- t}

\frac{d ^{2} y}{d x ^{2}} - 4 y = 4 sin (2 x) - 3 e^{2 x}

y^{^{''}} + y = cos (a t)

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = - 8 e^{x} + 6 e^{- x}

What is the solution for y^{''}-4y^'+3y=96e^{-5t}-24e^{-3t}+18 ?
The solution for y^{''}-4y^'+3y=96e^{-5t}-24e^{-3t}+18 is y=c_{1}e^{3t}+c_{2}e^t+2e^{-5t}-e^{-3t}+6