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(2/5 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{2}{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{2}{5} cos (2 t) + \frac{4}{5} sin (2 t)) + \frac{8}{5}

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

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

y^{^{''}} + y = 2 cos (3 x)

y^{^{''}} - 7 y^{^{'}} + 12 y = e^{4 x}

y^{^{''}} - y^{^{'}} + 49 y = 7 sin (7 t)

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 x + 3

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(2/5 cos(2t)+4/5 sin(2t))+8/5