What is the solution for y^{''}-2y^'+5y=-8,y(0)=2,y^'(0)=2 ?

The solution for y^{''}-2y^'+5y=-8,y(0)=2,y^'(0)=2 is y=e^t(18/5 cos(2t)-4/5 sin(2t))-8/5

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y^{''}-2y^'+5y=-8,y(0)=2,y^'(0)=2

y^{^{''}} - 2 y^{^{'}} + 5 y = - 8, y (0) = 2, y^{^{'}} (0) = 2

y = e^{t} (\frac{18}{5} cos (2 t) - \frac{4}{5} sin (2 t)) - \frac{8}{5}

Solution steps

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

Solve linear ODE:

y = e^{t} (\frac{18}{5} cos (2 t) - \frac{4}{5} sin (2 t)) - \frac{8}{5}

y = e^{t} (\frac{18}{5} cos (2 t) - \frac{4}{5} sin (2 t)) - \frac{8}{5}

\frac{1}{2} y^{^{''}} + 2 y = 2 tan (2 t) - \frac{1}{2} e^{t}

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

y^{^{''}} - y = x^{2} e^{x} + 9

y^{^{''}} - 2 y^{^{'}} - 3 y = 24 t e^{2 t}

y^{^{''}} - y^{^{'}} - 6 y = - 468 sin (3 t)

What is the solution for y^{''}-2y^'+5y=-8,y(0)=2,y^'(0)=2 ?
The solution for y^{''}-2y^'+5y=-8,y(0)=2,y^'(0)=2 is y=e^t(18/5 cos(2t)-4/5 sin(2t))-8/5