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

The solution for y^{''}-2y^'+5y=cos(x) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))-1/10 sin(x)+1/5 cos(x)

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y^{''}-2y^'+5y=cos(x)

y^{^{''}} - 2 y^{^{'}} + 5 y = cos (x)

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) - \frac{1}{10} sin (x) + \frac{1}{5} cos (x)

Solution steps

y^{''} (x) - 2 y^{'} (x) + 5 y = cos (x)

Solve linear ODE:

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) - \frac{1}{10} sin (x) + \frac{1}{5} cos (x)

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) - \frac{1}{10} sin (x) + \frac{1}{5} cos (x)

y^{^{''}} + 3 y = 5 t^{4}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 10 y^{^{'}} + 24 = 0

y^{^{'''}} + 3 y^{^{''}} - 4 = 0

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

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 e^{- 4 t}, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}-2y^'+5y=cos(x) ?
The solution for y^{''}-2y^'+5y=cos(x) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))-1/10 sin(x)+1/5 cos(x)