What is the solution for y^{''}-4y^'+5y=24cos(x) ?

The solution for y^{''}-4y^'+5y=24cos(x) is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))-3sin(x)+3cos(x)

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y^{''}-4y^'+5y=24cos(x)

y^{^{''}} - 4 y^{^{'}} + 5 y = 24 cos (x)

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) - 3 sin (x) + 3 cos (x)

Solution steps

y^{''} (x) - 4 y^{'} (x) + 5 y = 24 cos (x)

Solve linear ODE:

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) - 3 sin (x) + 3 cos (x)

y = e^{2 x} (c_{1} cos (x) + c_{2} sin (x)) - 3 sin (x) + 3 cos (x)

y^{^{'''}} + 3 y^{^{''}} - 4 = 5 t^{2} + 4

y^{^{''}} - y^{^{'}} = - 11

y^{^{''}} - y^{^{'}} = - 10

y^{^{'''}} - 16 y^{^{'}} = t e^{- 4 t} + 5 cos (4 t)

y^{^{''}} - 2 y^{^{'}} + 3 y = 3 sin (2 t)

What is the solution for y^{''}-4y^'+5y=24cos(x) ?
The solution for y^{''}-4y^'+5y=24cos(x) is y=e^{2x}(c_{1}cos(x)+c_{2}sin(x))-3sin(x)+3cos(x)