What is the solution for (D^2-1)y=e^{-x}(2sin(x)+4cos(x)) ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - y = x \cdot e^{2 x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 5 x + 4 - e^{5 x}

y^{^{''}} - 6 y^{^{'}} + 5 y = 34 sin (3 t)

y^{^{''}} - 9 y^{^{'}} + 2 y = t e^{t}

y^{^{''}} + 4 y = 8 sin (2 t) + 5 e^{t}

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