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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 10 y^{^{'}} + 24 y = e^{x} + 3 x

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

y^{^{''}} + y = cosh (x)

(D^{2} + D - 2) y = 2 x - 40 cos (2 x)

y^{^{''''}} + 16 y^{^{''}} = sin (4 t) + t e^{5 t} + 4

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