What is the solution for y^{''}+4y^'+5y=2cos(x)-2sin(x) ?

The solution for y^{''}+4y^'+5y=2cos(x)-2sin(x) is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/2 cos(x)

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y^{''}+4y^'+5y=2cos(x)-2sin(x)

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 cos (x) - 2 sin (x)

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = 2 cos (x) - 2 sin (x)

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} - 2 y = 2 x^{2} + 5

(D + 3) y = x + 1

y^{^{''}} - y = \frac{( 4 e ^{- x} )}{( 1 - e ^{- 2 x} )}

y^{^{''}} - 4 y^{^{'}} + 4 y = 10 x + 2

2 y^{^{''}} + y = 9 e^{2 x}

What is the solution for y^{''}+4y^'+5y=2cos(x)-2sin(x) ?
The solution for y^{''}+4y^'+5y=2cos(x)-2sin(x) is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/2 cos(x)