Question

If $\sin (x+y)\frac{dy}{dx}=5$, then.

• A. $\frac{dt}{1+5cosect}=dx$ (where $t=x+y$)
• B. $\frac{dt}{5\sin t+1}=dx$(where $t=x+y$)
• C. $5\int \frac{dt}{5+\sin t}=t-x$(where $t=x+y$)
• D. $5\int \frac{dt}{5+\sin t}=t+x$(where $t=x+y$)

Answer 1

Answer: (A)

$\frac{dt}{1+5cosect}=dx$ (where $t=x+y$)

Given differential equation is $\sin (x+y)\frac{dy}{dx}=5$,To solve this we take $x+y=t$,

$1+\frac{dy}{dx}=\frac{dt}{dx}$

$⟹\frac{dt}{dx}-1=5\csc t$,

$⟹\frac{1}{5\csc t+1}dt=dx$ where $t=x+y$