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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

u^{''} - u^{'} + 4 H (t (x)) = e^{5 t (x)}

y^{^{''}} - 2 y^{^{'}} + 17 y = e^{x} sin (x)

y^{^{''}} - 4 y^{^{'}} - 5 y = 2 + e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

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

4 x^{^{''}} + x = t^{2} + 2 cos (3 t)

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