What is the solution for y^{''}-3y^'-4y=4t^2-1 ?

The solution for y^{''}-3y^'-4y=4t^2-1 is y=c_{1}e^{4t}+c_{2}e^{-t}-t^2+(3t)/2-11/8

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y^{''}-3y^'-4y=4t^2-1

y^{^{''}} - 3 y^{^{'}} - 4 y = 4 t^{2} - 1

y = c_{1} e^{4 t} + c_{2} e^{- t} - t^{2} + \frac{3 t}{2} - \frac{11}{8}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = 4 t^{2} - 1

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- t} - t^{2} + \frac{3 t}{2} - \frac{11}{8}

y = c_{1} e^{4 t} + c_{2} e^{- t} - t^{2} + \frac{3 t}{2} - \frac{11}{8}

y^{^{''}} + 11 y^{^{'}} + 24 y = 4 sin (4 t) + 2 e^{- 8 t}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 2 y = 2 e^{x} cos (x)

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

y^{^{''}} + y = 2 sin (2 t), y (0) = 12, y^{^{'}} (0) = 0

10 y^{^{''}} + 90 y^{^{'}} + \frac{1}{0.05} y = 500 sin (x)

What is the solution for y^{''}-3y^'-4y=4t^2-1 ?
The solution for y^{''}-3y^'-4y=4t^2-1 is y=c_{1}e^{4t}+c_{2}e^{-t}-t^2+(3t)/2-11/8