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

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

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - \frac{4}{25} sin (x) + \frac{3}{25} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - \frac{4}{25} sin (x) + \frac{3}{25} cos (x)

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - \frac{4}{25} sin (x) + \frac{3}{25} cos (x)

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

(D + 2) (D - 1)^{3} y = e^{x}

y^{^{''}} + 16 y = 5 sin (4 x)

y^{^{''}} + 9 y = cos (x), y (\frac{π}{2}) = - \frac{5}{3}, y^{^{'}} (0) = 5

(D^{2} - 3 D + 2) y = x + 1

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