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

The solution for y^{''}-y^'-2y=3t^2-1 is y=c_{1}e^{2t}+c_{2}e^{-t}-(3t^2)/2+(3t)/2-7/4

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

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

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{3 t ^{2}}{2} + \frac{3 t}{2} - \frac{7}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{3 t ^{2}}{2} + \frac{3 t}{2} - \frac{7}{4}

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{3 t ^{2}}{2} + \frac{3 t}{2} - \frac{7}{4}

y^{^{''}} + 36 y = 33 cos (5 x)

y^{^{'''}} - y^{^{'}} = 2 \cdot cos (t)

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

y^{^{''}} - 3 y^{^{'}} - 6 y = 2 t^{2} + 2 t - 5 e^{3 t}

d (x) = 2 x - 4

What is the solution for y^{''}-y^'-2y=3t^2-1 ?
The solution for y^{''}-y^'-2y=3t^2-1 is y=c_{1}e^{2t}+c_{2}e^{-t}-(3t^2)/2+(3t)/2-7/4