We are given to solve the first order linear ODE:
$\displaystyle \frac{df}{dy}+f(y)=\sin(2y)$
I am assuming $\displaystyle f$ is the dependent variable and $\displaystyle y$ is the independent variable, and that the (t) is a typo.
We may begin by calculating our integrating factor $\displaystyle \mu(y)$:
$\displaystyle \mu(y)=e^{\int\,dy}=e^y$
Multiply the ODE by the integrating factor:
$\displaystyle e^y\frac{df}{dy}+f(y)e^y=e^y\sin(2y)$
Now, we may rewrite the left side as the differentiation of a product:
$\displaystyle \frac{d}{dy}\left(e^yf \right)=e^y\sin(2y)$
Integrate with respect to $\displaystyle y$:
$\displaystyle \int\frac{d}{dy}\left(e^yf \right)\,dy=\int e^y\sin(2y)\,dy$
On the right side, we may use integration by parts:
$\displaystyle u=\sin(2y)\,\therefore\,du=2\cos(2y)\,dy$
$\displaystyle dv=e^y\,dy\,\therefore\,v=e^y$
and so we may state:
$\displaystyle I=\int e^y\sin(2y)\,dy=e^y\sin(2y)-2\int e^y\cos(2y)\,dy$
Now, using integration by parts again:
$\displaystyle u=\cos(2y)\,\therefore\,du=-2\sin(2y)\,dy$
$\displaystyle dv=e^y\,dy\,\therefore\,v=e^y$
and we have:
$\displaystyle I=e^y\sin(2y)-2\left(e^y\cos(2y)+2\int e^y\sin(2y)\,dy \right)$
$\displaystyle I=e^y\sin(2y)-2e^y\cos(2y)-4I$
Solve for $\displaystyle I$:
$\displaystyle I=\frac{e^y(\sin(2y)-2\cos(2y))}{5}+c_1$
Now, back to integrating the ODE, we have:
$\displaystyle e^yf=\frac{e^y(\sin(2y)-2\cos(2y))}{5}+c_1$
Hence, the general solution is given by:
$\displaystyle f(y)=\frac{\sin(2y)-2\cos(2y)}{5}+c_1e^{-y}$