What is the solution for y^{''}-2y^'-3y=3e^{5t} ?

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

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y^{''}-2y^'-3y=3e^{5t}

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 e^{5 t}

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 3 e^{5 t}

Solve linear ODE:

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

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

y^{^{''}} + 36 y = 8 cos (6 x) + 12 sin (6 x)

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 5 y = 24 e^{x}

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 3 y = 6 x^{2} + 19 x + 5

y^{^{'''}} + y^{^{''}} - 2 y = x e^{x} + 5

(D^{4} - 8 D^{2} + 16) y = x e^{2 x}

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