$\frac{dy}{dx} = 3\sin(x/2)$
$y = 1 ,x= \frac{\pi}{3}$
I just get stuck on the integration of the $3\sin(x/2)$.
$\frac{dy}{dx} = 3\sin(x/2)$
$y = 1 ,x= \frac{\pi}{3}$
I just get stuck on the integration of the $3\sin(x/2)$.
$\int\sin kx\,dx=-\frac{1}{k}\cos kx+C\,\,,\,k\neq 0\,$
Begin by getting the general solution, $y = \int 3\sin(x/2)\, dx $ You can then use the second condition to solve for the constant of integration and obtain a solution to the DE.
Here $\displaystyle \frac{dy}{dx}=3\sin \frac{x}{2}$.
$\displaystyle \Rightarrow dy= 3\sin \frac{x}{2} dx$ (variable saperable)
$\displaystyle \Rightarrow \int dy=3\int \sin \frac{x}{2} dx+ C$
$\displaystyle \therefore y=-3 \frac{\cos{\frac{x}{2}}}{\frac{1}{2}}+C$.
i.e. $\displaystyle y=-6\cos \frac{x}{2}+C$.
Now, $\displaystyle y=1$ and $\displaystyle x=\frac{\pi}{3}$, we get
$\displaystyle 1=-6\cos \frac{\pi}{6} +C$ $\Rightarrow 1=-6 \frac{\sqrt{3}}{2}+C$ $\Rightarrow C=1+3\sqrt{3}.$
The particular solution of give differential equation is $\displaystyle y=-6\cos \frac{x}{2}+1+3\sqrt{3}$.