What is the solution for (d^2+4d+5)*y=-2cos(x) ?

The solution for (d^2+4d+5)*y=-2cos(x) is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-1/4 sin(x)-1/4 cos(x)

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(d^2+4d+5)*y=-2cos(x)

(d^{2} + 4 d + 5) \cdot y = - 2 cos (x)

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

Solution steps

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

Solve linear ODE:

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

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

(d^{3} - 3 d + 2) \cdot y = 540 x^{3} \cdot e^{- x}

(d^{3} + 8) \cdot y = x^{4} + 2 x + 1 + 3 sin (x)

(d^{2} + d - 2) y = 1 + x - x^{2}

(d^{3} - d^{2} - d + 1) \cdot y = x (e^{- x} + 1)

(d^{3} + 2 d^{2} - 3 d) \cdot y = x \cdot e^{3 x}

What is the solution for (d^2+4d+5)*y=-2cos(x) ?
The solution for (d^2+4d+5)*y=-2cos(x) is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))-1/4 sin(x)-1/4 cos(x)