Search any question & find its solution
Question:
Answered & Verified by Expert
The solution of the differential equation $\frac{d y}{d x}=1-\cos (y-x) \cot (y-x)$ is
Options:
Solution:
1192 Upvotes
Verified Answer
The correct answer is:
$x+\sec (y-x)=c$
We have,
$$
\frac{d y}{d x}=1-\cos (y-x) \cot (y-x)
$$
Put $y-x=v \Rightarrow \frac{d y}{d x}=1+\frac{d v}{d x}$
$$
\begin{aligned}
& \therefore 1+\frac{d v}{d x}=1-\cos v \cot v \Rightarrow \frac{d v}{d x}=-\frac{\cos ^2 v}{\sin v} \\
& \Rightarrow \quad-\int \frac{\sin v}{\cos ^2 v} d v=\int d x \Rightarrow-\int \sec v \tan v d v=d x \\
& \Rightarrow \quad-\sec v=x+c \Rightarrow x+\sec (y-x)=c
\end{aligned}
$$
$$
\frac{d y}{d x}=1-\cos (y-x) \cot (y-x)
$$
Put $y-x=v \Rightarrow \frac{d y}{d x}=1+\frac{d v}{d x}$
$$
\begin{aligned}
& \therefore 1+\frac{d v}{d x}=1-\cos v \cot v \Rightarrow \frac{d v}{d x}=-\frac{\cos ^2 v}{\sin v} \\
& \Rightarrow \quad-\int \frac{\sin v}{\cos ^2 v} d v=\int d x \Rightarrow-\int \sec v \tan v d v=d x \\
& \Rightarrow \quad-\sec v=x+c \Rightarrow x+\sec (y-x)=c
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.