What is the solution for y^{''}-4y^'-32y=108e^{2t} ?

The solution for y^{''}-4y^'-32y=108e^{2t} is y=c_{1}e^{8t}+c_{2}e^{-4t}-3e^{2t}

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y^{''}-4y^'-32y=108e^{2t}

y^{^{''}} - 4 y^{^{'}} - 32 y = 108 e^{2 t}

y = c_{1} e^{8 t} + c_{2} e^{- 4 t} - 3 e^{2 t}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 32 y = 108 e^{2 t}

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} e^{- 4 t} - 3 e^{2 t}

y = c_{1} e^{8 t} + c_{2} e^{- 4 t} - 3 e^{2 t}

y^{^{''}} + 16 y = 32 sec (2 x)

y^{^{''''}} - y^{^{'''}} - y^{^{''}} - y^{^{'}} - 2 y = 8 x^{5}

(d^{2} - 2 d + 5) y = e^{x}

(d^{3} - d) y = 2 cosh (x)

(d^{2} + 1) \cdot y = x \cdot cos (x)

What is the solution for y^{''}-4y^'-32y=108e^{2t} ?
The solution for y^{''}-4y^'-32y=108e^{2t} is y=c_{1}e^{8t}+c_{2}e^{-4t}-3e^{2t}