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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int x^{3} sin^{2} (1 + x^{\frac{5}{2}}) d x

\int_{- 1}^{1} x^{2} - 1 d x

\int_{0}^{0.5} 750 x d x

2 e^{x} cos (x)

y^{^{'}} + (\frac{1}{x} - 1) y = - \frac{2}{x}

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