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

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

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

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

y = c_{1} e^{2 t} + c_{2} e^{- 4 t} - \frac{t}{4} - \frac{1}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 4 t} - \frac{t}{4} - \frac{1}{16}

y = c_{1} e^{2 t} + c_{2} e^{- 4 t} - \frac{t}{4} - \frac{1}{16}

y^{^{''}} + y = - 16 \cdot cos (3 x)

y^{^{''}} - y = \frac{e ^{x}}{x ^{2}}

y^{^{''}} - 9 y = 1, y^{^{'}} (0) = 10

y^{^{''}} + y = 9 e^{- 4 x}, y (0) = 0, y^{^{'}} (0) = 0

x^{^{''}} + 2 x^{^{'}} + 5 x = 10 cos (3 t)

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