What is the solution for y^{''}-2y^'+5y=-8 ?

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

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y^{''}-2y^'+5y=-8

y^{^{''}} - 2 y^{^{'}} + 5 y = - 8

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) - \frac{8}{5}

Solution steps

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

Solve linear ODE:

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) - \frac{8}{5}

y = e^{t} (c_{1} cos (2 t) + c_{2} sin (2 t)) - \frac{8}{5}

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

\frac{d ^{2} x}{d t ^{2}} + 81 x = 2 sin (9 t)

(D^{2} - 1) y = e^{- x} (2 sin (x) + 4 cos (x))

\frac{d ^{2} y}{d x ^{2}} - y = x \cdot e^{2 x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 5 x + 4 - e^{5 x}

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