What is the solution for y^{''}-2y^'-8y=5cos(t) ?

The solution for y^{''}-2y^'-8y=5cos(t) is y=c_{1}e^{4t}+c_{2}e^{-2t}-2/17 sin(t)-9/17 cos(t)

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y^{''}-2y^'-8y=5cos(t)

y^{^{''}} - 2 y^{^{'}} - 8 y = 5 cos (t)

y = c_{1} e^{4 t} + c_{2} e^{- 2 t} - \frac{2}{17} sin (t) - \frac{9}{17} cos (t)

Solution steps

y^{''} (t) - 2 y^{'} (t) - 8 y = 5 cos (t)

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- 2 t} - \frac{2}{17} sin (t) - \frac{9}{17} cos (t)

y = c_{1} e^{4 t} + c_{2} e^{- 2 t} - \frac{2}{17} sin (t) - \frac{9}{17} cos (t)

y^{^{''}} + 2 y^{^{'}} - 8 y = 8 x^{2} - 3

\frac{d ^{2} y}{d x ^{2}} + y = - 16 \cdot cos (3 x)

(D - 2)^{2} y = 20 - 3 x e^{2} x

y^{^{''}} - y^{^{'}} - 15 = 0

y^{^{''}} - y^{^{'}} + 3 y = 9 t + 2

What is the solution for y^{''}-2y^'-8y=5cos(t) ?
The solution for y^{''}-2y^'-8y=5cos(t) is y=c_{1}e^{4t}+c_{2}e^{-2t}-2/17 sin(t)-9/17 cos(t)