What is the solution for y^{''}+4y=sin(2x)+1 ?

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

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y^{''}+4y=sin(2x)+1

y^{^{''}} + 4 y = sin (2 x) + 1

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{- x cos ( 2 x ) + 1}{4}

Solution steps

y^{''} (x) + 4 y = sin (2 x) + 1

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{- x cos ( 2 x ) + 1}{4}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{- x cos ( 2 x ) + 1}{4}

x^{^{''}} + 8 x^{^{'}} + 16 x = 6 sin (4 t)

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

x^{^{''}} + 16 x = 2 sin (t)

y^{^{''}} - 6 y^{^{'}} + 5 y = 9 e^{2 x}

y^{^{'''}} + y^{^{'}} = - 2 sin (t)

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