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

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

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(dx)/(dt)=sin(2t)-3x

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial y} (sin (y) - x y + y^{2})

\int \frac{1}{( x + 2 ) x ^{2} + 4 x + 3} d x

\int 2 sin^{3} (x co) s^{2} x d x

3 x^{2}, 6 x

\frac{d}{d x} (ln (3 x^{2} - 5 x + 2))

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