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

The solution for y^{''}-3y^'+2y=e^xsin(2x) is y=c_{1}e^{2x}+c_{2}e^x+1/10 e^xcos(2x)-1/5 e^xsin(2x)

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

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

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{10} e^{x} cos (2 x) - \frac{1}{5} e^{x} sin (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{10} e^{x} cos (2 x) - \frac{1}{5} e^{x} sin (2 x)

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{1}{10} e^{x} cos (2 x) - \frac{1}{5} e^{x} sin (2 x)

\frac{\partial}{\partial x} (arctan (2 x y))

\int \frac{( e ^{x} - e ^{- x} )}{e ^{2 x}} d x

\frac{\partial}{\partial s} (s cos (t))

\frac{d}{d x} (x^{2} - 4 - 2 arcsec (\frac{x}{2}))

What is the solution for y^{''}-3y^'+2y=e^xsin(2x) ?
The solution for y^{''}-3y^'+2y=e^xsin(2x) is y=c_{1}e^{2x}+c_{2}e^x+1/10 e^xcos(2x)-1/5 e^xsin(2x)