What is the solution for (dy)/(dt)=sin(3t)-2y ?

The solution for (dy)/(dt)=sin(3t)-2y is y=-(3cos(3t))/(13)+(2sin(3t))/(13)+(c_{1})/(e^{2t)}

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(dy)/(dt)=sin(3t)-2y

\frac{d y}{d t} = sin (3 t) - 2 y

y = - \frac{3 cos ( 3 t )}{13} + \frac{2 sin ( 3 t )}{13} + \frac{c _{1}}{e ^{2 t}}

Solution steps

\frac{d y}{d t} = sin (3 t) - 2 y

Solve linear ODE:

y = - \frac{3 cos ( 3 t )}{13} + \frac{2 sin ( 3 t )}{13} + \frac{c _{1}}{e ^{2 t}}

y = - \frac{3 cos ( 3 t )}{13} + \frac{2 sin ( 3 t )}{13} + \frac{c _{1}}{e ^{2 t}}

x \to 9 + lim (\frac{5}{x - 9})

\int (5 + 3 x + 1) d x

3 sec (x)

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

\frac{d}{d x} (3 + ln (x))

What is the solution for (dy)/(dt)=sin(3t)-2y ?
The solution for (dy)/(dt)=sin(3t)-2y is y=-(3cos(3t))/(13)+(2sin(3t))/(13)+(c_{1})/(e^{2t)}