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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = tan (t) + e^{3 t} - 1

\frac{\partial}{\partial u} (\frac{u}{v})

\frac{\partial}{\partial y} (y^{2} - 2 y cos (x))

ln (x e^{7 x})

f (x) = \frac{32}{x}

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