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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 17 y^{^{'}} + 72 y = 55 + 72 x

y^{^{''}} - y^{^{'}} - 6 y = 5 e^{2 x}

y^{^{''}} + 2 y^{^{'}} + 7 y = 6 cos (2 x)

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

(D^{2} - 4 D + 3) y = e^{3 x} cos (2 x)

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