What is the solution for y^{''}+4y=sin(2x+3) ?

The solution for y^{''}+4y=sin(2x+3) is y=c_{1}cos(2x)+c_{2}sin(2x)-(xcos(2x+3))/4

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y^{''}+4y=sin(2x+3)

y^{^{''}} + 4 y = sin (2 x + 3)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x cos ( 2 x + 3 )}{4}

Solution steps

y^{''} (x) + 4 y = sin (2 x + 3)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x cos ( 2 x + 3 )}{4}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x cos ( 2 x + 3 )}{4}

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

y^{^{''}} + y^{^{'}} - 12 y = - 40 t + 24 t^{2}

y^{^{''}} + y^{^{'}} + 4 y = e^{t} + e^{- t}

(d^{2} + 2 d + 2) y = 3 e^{x} sin (x)

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

What is the solution for y^{''}+4y=sin(2x+3) ?
The solution for y^{''}+4y=sin(2x+3) is y=c_{1}cos(2x)+c_{2}sin(2x)-(xcos(2x+3))/4