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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = 8 cos (2 x) - 4 sin (x)

y^{^{''}} - 4 y = sin^{2} (x)

y^{^{''}} - y = e^{2 x}, y (0) = 1, y^{^{'}} (0) = 0

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

y^{^{''}} + y = 7 x sin (x)

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