What is the solution for (D^2+2D+1)y=4sin(2x) ?

The solution for (D^2+2D+1)y=4sin(2x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-12/25 sin(2x)-16/25 cos(2x)

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(D^2+2D+1)y=4sin(2x)

(D^{2} + 2 D + 1) y = 4 sin (2 x)

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{12}{25} sin (2 x) - \frac{16}{25} cos (2 x)

Solution steps

(D^{2} + 2 D + 1) y = 4 sin (2 x)

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{12}{25} sin (2 x) - \frac{16}{25} cos (2 x)

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{12}{25} sin (2 x) - \frac{16}{25} cos (2 x)

x^{^{''}} - 16 x^{^{'}} + 64 x = 3 t e^{8 t}

(D^{3} + 4 D) y = e^{x} + sin (x)

y^{^{''}} + y = - cos (2 t), y^{^{'}} (0) = 0, y (0) = 0

y^{^{''}} - 6 y^{^{'}} + 5 y = 12 e^{t}, y (0) = - 1, y^{^{'}} (0) = 7

y^{^{''}} + 9 y = 4 + 5 x

What is the solution for (D^2+2D+1)y=4sin(2x) ?
The solution for (D^2+2D+1)y=4sin(2x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-12/25 sin(2x)-16/25 cos(2x)