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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = \frac{x - 7}{9 x ^{2} e ^{x}}

\frac{\partial}{\partial x} ((e^{x} - 3) sin (y))

\frac{d}{d x} ((3 x + 2) (4 x - 3))

\frac{d}{d x} ((x^{3} - 4 x) \cdot e^{3 x})

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