What is the solution for (D^2+1)y=sin(x) ?

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

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(D^2+1)y=sin(x)

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = 16 cos (3 x)

y^{^{''}} + 7 y^{^{'}} + 12 y = e^{t}

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 5 y = e^{- x}

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

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{8 e ^{- 3 x}}{x ^{2} + 1}

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