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

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

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(D^2+D)y=-cos(x)

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

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

Solution steps

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

Solve linear ODE:

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

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

4 y^{^{''}} + 2 y^{^{'}} + 70 y = 4 cos (4 t)

y^{^{''''}} - 2 y^{^{''}} - 8 y = 6 e^{2 x}

y^{^{''}} - 5 y^{^{'}} + 6 y = π x^{2} + 3

y^{^{'''}} + y^{^{''}} = 9 x^{2}

y^{^{''}} + 2 y^{^{'}} - 3 y = 7 cos (3 x)

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