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

The solution for y^{''}-2y^'-3y=2sin(x) is y=c_{1}e^{3x}+c_{2}e^{-x}-2/5 sin(x)+1/5 cos(x)

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

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

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{2}{5} sin (x) + \frac{1}{5} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{2}{5} sin (x) + \frac{1}{5} cos (x)

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{2}{5} sin (x) + \frac{1}{5} cos (x)

y^{^{''}} - 2 y^{^{'}} + y = e^{3 x} + e^{2 x}

(D^{2} - 4 D + 4) y = sin (3 x)

y^{^{''}} - 2 y^{^{'}} - 15 = 0, y (0) = 2, y^{^{'}} (0) = 6

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{3 x} - x^{2}

\frac{d ^{2} y}{d x ^{2}} + 4 y = - 3 x^{2} + 2 x + 3

What is the solution for y^{''}-2y^'-3y=2sin(x) ?
The solution for y^{''}-2y^'-3y=2sin(x) is y=c_{1}e^{3x}+c_{2}e^{-x}-2/5 sin(x)+1/5 cos(x)