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

The solution for y^'+y=sin(2t) is y=-(2cos(2t))/5+(sin(2t))/5+(c_{1})/(e^t)

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y^'+y=sin(2t)

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

y = - \frac{2 cos ( 2 t )}{5} + \frac{sin ( 2 t )}{5} + \frac{c _{1}}{e ^{t}}

Solution steps

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

Solve linear ODE:

y = - \frac{2 cos ( 2 t )}{5} + \frac{sin ( 2 t )}{5} + \frac{c _{1}}{e ^{t}}

y = - \frac{2 cos ( 2 t )}{5} + \frac{sin ( 2 t )}{5} + \frac{c _{1}}{e ^{t}}

\frac{\partial}{\partial y} (x y + yz)

x \to - 1 - lim (\frac{- 1}{x ^{2} - 1})

1 + x^{3}

\frac{d}{d x} (e^{r (x) x})

\frac{d}{d y} (sin (x) cos (y))

What is the solution for y^'+y=sin(2t) ?
The solution for y^'+y=sin(2t) is y=-(2cos(2t))/5+(sin(2t))/5+(c_{1})/(e^t)