What is the solution for y^{''}-y^'-2y=3sin(x) ?

The solution for y^{''}-y^'-2y=3sin(x) is y=c_{1}e^{2x}+c_{2}e^{-x}-9/10 sin(x)+3/10 cos(x)

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y^{''}-y^'-2y=3sin(x)

y^{^{''}} - y^{^{'}} - 2 y = 3 sin (x)

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{9}{10} sin (x) + \frac{3}{10} cos (x)

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 3 sin (x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{9}{10} sin (x) + \frac{3}{10} cos (x)

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{9}{10} sin (x) + \frac{3}{10} cos (x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 36 t + 6

y^{^{''}} + y^{^{'}} = 3 e^{\frac{x}{2}}

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 e^{5 t}

y^{^{''}} + 36 y = 8 cos (6 x) + 12 sin (6 x)

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 5 y = 24 e^{x}

What is the solution for y^{''}-y^'-2y=3sin(x) ?
The solution for y^{''}-y^'-2y=3sin(x) is y=c_{1}e^{2x}+c_{2}e^{-x}-9/10 sin(x)+3/10 cos(x)