What is the solution for y^{''}-2y^'-8y=-16t+16t^2 ?

The solution for y^{''}-2y^'-8y=-16t+16t^2 is y=c_{1}e^{4t}+c_{2}e^{-2t}-2t^2+3t-5/4

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y^{''}-2y^'-8y=-16t+16t^2

y^{^{''}} - 2 y^{^{'}} - 8 y = - 16 t + 16 t^{2}

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 8 y = - 16 t + 16 t^{2}

Solve linear ODE:

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

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

y^{^{''}} - 16 y = 48 t - 24 e^{- 4 t}

x^{^{''}} + 2 x^{^{'}} + 50 x = 2 cos (7 t)

y^{^{''}} + y = (cos (x))^{2}

\frac{d ^{2} y}{d t ^{2}} + 4 y = 4 sec^{2} (2 t)

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

What is the solution for y^{''}-2y^'-8y=-16t+16t^2 ?
The solution for y^{''}-2y^'-8y=-16t+16t^2 is y=c_{1}e^{4t}+c_{2}e^{-2t}-2t^2+3t-5/4