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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 36 y = sec (6 x)

y^{^{''}} + 16 y = t^{2} + 1, y (0) = 0, y^{^{'}} (0) = 1

y^{^{'''}} - 12 y^{^{''}} + 48 y^{^{'}} - 64 y = 9 e^{4 x}

y^{^{'''}} + 121 y^{^{'}} = tan (11 x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 x e^{2 x} + 4 x - 5

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