What is the solution for y^{''}+2y^'=49e^xsin(2x) ?

The solution for y^{''}+2y^'=49e^xsin(2x) is y=c_{1}+c_{2}e^{-2x}-49/65 e^xsin(2x)-392/65 e^xcos(2x)

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y^{''}+2y^'=49e^xsin(2x)

y^{^{''}} + 2 y^{^{'}} = 49 e^{x} sin (2 x)

y = c_{1} + c_{2} e^{- 2 x} - \frac{49}{65} e^{x} sin (2 x) - \frac{392}{65} e^{x} cos (2 x)

Solution steps

y^{''} (x) + 2 y^{'} (x) = 49 e^{x} sin (2 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} - \frac{49}{65} e^{x} sin (2 x) - \frac{392}{65} e^{x} cos (2 x)

y = c_{1} + c_{2} e^{- 2 x} - \frac{49}{65} e^{x} sin (2 x) - \frac{392}{65} e^{x} cos (2 x)

y^{^{''}} + 2 y^{^{'}} + 1 = t e^{t} sin (t)

y^{^{''''}} + 2 y^{^{''}} + y = 3 t + 10

y^{^{''}} - 5 y^{^{'}} + 6 y = \frac{e ^{2 x}}{1 + e ^{x}}

x^{^{''}} + 3 x^{^{'}} + 2 x = cos (t)

y^{^{''}} + 2 y = cos (x)

What is the solution for y^{''}+2y^'=49e^xsin(2x) ?
The solution for y^{''}+2y^'=49e^xsin(2x) is y=c_{1}+c_{2}e^{-2x}-49/65 e^xsin(2x)-392/65 e^xcos(2x)