What is the solution for y^{''}+2y^'-8y=4a ?

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

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y^{''}+2y^'-8y=4a

y^{^{''}} + 2 y^{^{'}} - 8 y = 4 a

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

Solution steps

y^{''} (t) + 2 y^{'} (t) - 8 y = 4 a

Solve linear ODE:

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

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

(D^{2} + 4 D + 8) y = 4 sin (2 x)

y^{^{''}} + 9 y = 2 t^{2} - 2, y (0) = 0, y^{^{'}} (0) = - 8

y^{^{'''}} + 3 y^{^{''}} - y^{^{'}} - 3 y = 3 x - 14

y^{^{''}} - 16 y^{^{'}} + 64 y = - \frac{2}{3 x} e^{8 x}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = 4 (x - 1)

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