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

The solution for y^{''}-3y^'-4y=t^2 is y=c_{1}e^{4t}+c_{2}e^{-t}-(t^2)/4+(3t)/8-13/32

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

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

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{t ^{2}}{4} + \frac{3 t}{8} - \frac{13}{32}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{t ^{2}}{4} + \frac{3 t}{8} - \frac{13}{32}

y = c_{1} e^{4 t} + c_{2} e^{- t} - \frac{t ^{2}}{4} + \frac{3 t}{8} - \frac{13}{32}

y^{^{''}} + y = 5 cos (x), y (0) = 0, y (0) = 1

y^{^{''}} - 2 y^{^{'}} + y = 2 e^{t}, y (0) = 1, y^{^{'}} (0) = 1

x^{^{''}} - x^{^{'}} - 6 x = 5 t

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 7 e x

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{- t}

What is the solution for y^{''}-3y^'-4y=t^2 ?
The solution for y^{''}-3y^'-4y=t^2 is y=c_{1}e^{4t}+c_{2}e^{-t}-(t^2)/4+(3t)/8-13/32