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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

n = 0 \sum \infty \frac{n}{2 n}

h \to 0 lim (\frac{h}{sin ( 3 h )})

\frac{\partial}{\partial x} (4 x^{2} - 2 y)

\frac{d}{d x} (ln ((sin (x) cos (x))^{2}))

\int x^{2} in (x) d x

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