What is the solution for y^'+3y=2sin(3t) ?

The solution for y^'+3y=2sin(3t) is y=(-cos(3t)+sin(3t))/3+(c_{1})/(e^{3t)}

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y^'+3y=2sin(3t)

y^{^{'}} + 3 y = 2 sin (3 t)

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

Solution steps

y^{'} (t) + 3 y = 2 sin (3 t)

Solve linear ODE:

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

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

\frac{x ^{2} - 35}{x - 6}

- 16 sin (2 x)

\int sec (1 - x) tan (1 - x) d x

\int (- \frac{4}{x ^{3}} - \frac{8}{x ^{5}}) d x

\int \frac{x}{x + x ^{\frac{4}{5}}} d x

What is the solution for y^'+3y=2sin(3t) ?
The solution for y^'+3y=2sin(3t) is y=(-cos(3t)+sin(3t))/3+(c_{1})/(e^{3t)}