Question

Solve differential equation $\frac{dy}{dx}=\sin (x+y)$

Answer 1

$\therefore \frac{dt}{dx}-1=\sin t(takex+y=t)$
$\therefore \frac{dt}{dx}=1+\sin t$
$\therefore \frac{dt}{1+\sin t}=dx$
Integrating on both the sides
$\therefore \int \frac{1-\sin t}{1-{\sin }^{2}t}dt=\int dx$
$\therefore \int \frac{1}{{\cos }^{2}t}dt-\int \frac{\sin t}{{\cos }^{2}t}dt=\int dx$
$\therefore \int {\sec }^{2}tdt-\int \tan t\sec tdt=\int dx$
$\therefore \tan t-\sec t=x+c$
$\therefore \tan (x+y)-\sec (x+y)=x+c$