What is the solution for y^{''}+2y^'+2y=10cos(x) ?

The solution for y^{''}+2y^'+2y=10cos(x) is y=e^{-x}(c_{1}cos(x)+c_{2}sin(x))+4sin(x)+2cos(x)

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y^{''}+2y^'+2y=10cos(x)

y^{^{''}} + 2 y^{^{'}} + 2 y = 10 cos (x)

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + 2 y = 10 cos (x)

Solve linear ODE:

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

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

(D^{2} - 1) y = x^{2} e^{\frac{x ^{2}}{2}}

d, d (x) = \frac{k ( x ) + 18 x - 9}{2 x ^{2}}

y^{^{'''}} + 3 y^{^{''}} - 4 y = x e^{x} + 14 x

y^{^{''}} - 2 y^{^{'}} = - 6 x^{2} + 1

y^{^{''}} + y^{^{'}} + 0.25 y = e^{x} (sin (3 x) - cos (3 x))

What is the solution for y^{''}+2y^'+2y=10cos(x) ?
The solution for y^{''}+2y^'+2y=10cos(x) is y=e^{-x}(c_{1}cos(x)+c_{2}sin(x))+4sin(x)+2cos(x)