What is the solution for (d^2+4d+4)y=10e^{-x}cos(x) ?

The solution for (d^2+4d+4)y=10e^{-x}cos(x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+5e^{-x}sin(x)

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(d^2+4d+4)y=10e^{-x}cos(x)

(d^{2} + 4 d + 4) y = 10 e^{- x} cos (x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + 5 e^{- x} sin (x)

Solution steps

(d^{2} + 4 d + 4) y = 10 e^{- x} cos (x)

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + 5 e^{- x} sin (x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + 5 e^{- x} sin (x)

(d^{2} + 5 d + 6) \cdot y = e^{x} + e^{- x}

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

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

(d^{2} + 4) \cdot y = 4 tan^{2} (x)

y^{^{''}} - 3 y^{^{'}} = x + 3

What is the solution for (d^2+4d+4)y=10e^{-x}cos(x) ?
The solution for (d^2+4d+4)y=10e^{-x}cos(x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+5e^{-x}sin(x)