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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} + 64 y = 5 cos (2 t)

y^{^{''''}} - y^{^{'''}} = x + e^{x}

y^{^{'''}} + 9 y^{^{'}} = tan (3 x)

y^{^{''}} - y^{^{'}} + 14 y = \frac{4 + e x}{2}

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

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