What is the solution for y^{''}-4y^'+4y=25cos(x) ?

The solution for y^{''}-4y^'+4y=25cos(x) is y=c_{1}e^{2x}+c_{2}xe^{2x}-4sin(x)+3cos(x)

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y^{''}-4y^'+4y=25cos(x)

y^{^{''}} - 4 y^{^{'}} + 4 y = 25 cos (x)

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - 4 sin (x) + 3 cos (x)

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 25 cos (x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - 4 sin (x) + 3 cos (x)

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - 4 sin (x) + 3 cos (x)

x^{^{''}} - x = e^{- t} sin (e^{- t}) + cos (e^{- t})

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e x - 10

y^{^{'''}} + 3 y^{^{''}} + y^{^{'}} - 5 y = - 5 x

y^{^{''}} + 4 y = 4 sin (2 t) + 8 cos (2 t)

y^{^{''}} + 4 y = 2 cos (x)

What is the solution for y^{''}-4y^'+4y=25cos(x) ?
The solution for y^{''}-4y^'+4y=25cos(x) is y=c_{1}e^{2x}+c_{2}xe^{2x}-4sin(x)+3cos(x)