The particular solution of the differential equation $\left(y+x \cdot \frac{d y}{d x}\right) \cdot \sin x…
The particular solution of the differential equation $\left(y+x \cdot \frac{d y}{d x}\right) \cdot \sin x y=\cos x$
at $x=0$ is
$\sin x+\cos x y=1$
$\cos x-\sin x y=1$
$\sin x-\cos x y=1$
$\cos x+\sin x y=1$
Solution
We have $\left(y+x \frac{d y}{d x}\right) \sin x y=\cos x$
Put $x y=u \Rightarrow x \frac{d y}{d x}+y=\frac{d u}{d x}$
$\therefore\left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \sin \mathrm{u}=\cos \mathrm{x}$
$\therefore \int \sin u d u=\int \cos x d x \Rightarrow-\cos u=\sin x+c \Rightarrow-\cos x y=\sin x+c$
When $x=0$, we get
$-\cos 0=0+c \Rightarrow c=-1$
$\therefore-\cos x y=\sin x-1 \Rightarrow \sin x+\cos x y=1$