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

The solution for 5y^{''}+y=sin(2x)+x is y=c_{1}cos(x/(sqrt(5)))+c_{2}sin(x/(sqrt(5)))-1/19 sin(2x)+x

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

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

y = c_{1} cos (\frac{x}{5}) + c_{2} sin (\frac{x}{5}) - \frac{1}{19} sin (2 x) + x

Solution steps

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

Solve linear ODE:

y = c_{1} cos (\frac{x}{5}) + c_{2} sin (\frac{x}{5}) - \frac{1}{19} sin (2 x) + x

y = c_{1} cos (\frac{x}{5}) + c_{2} sin (\frac{x}{5}) - \frac{1}{19} sin (2 x) + x

(4 D^{2} + 4 D + 1) y = x e^{- \frac{x}{2}} sin (x)

y^{^{''}} + 2 y^{^{'}} + 4 y = 111 e^{2 x} cos (3 x)

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

y^{^{''}} + 9 y = 5 sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 4

What is the solution for 5y^{''}+y=sin(2x)+x ?
The solution for 5y^{''}+y=sin(2x)+x is y=c_{1}cos(x/(sqrt(5)))+c_{2}sin(x/(sqrt(5)))-1/19 sin(2x)+x