Question

The solution of the differential equation
\[ \frac{dy}{dx}=\sin(x+y)\tan(x+y)-1 \] is

• A. \(\cosec(x+y)+\tan(x+y)=x+c\)
• B. \(x+\cosec(x+y)=c\)
• C. \(x+\tan(x+y)=c\)
• D. \(x+\sec(x+y)=c\)

Hint:

When equation contains \(x+y\), substitute \(u=x+y\). Then rewrite \(dy/dx\) using \(du/dx=1+dy/dx\).

Answer 1

The Correct Option is B

Step 1: Substitute \(u=x+y\).
\[ u=x+y \Rightarrow \frac{du}{dx}=1+\frac{dy}{dx} \]
Given:
\[ \frac{dy}{dx}=\sin u \tan u-1 \]
So:
\[ \frac{du}{dx}=1+\sin u\tan u-1=\sin u\tan u \]
Step 2: Simplify \(\sin u\tan u\).
\[ \sin u\tan u=\sin u\cdot\frac{\sin u}{\cos u}=\frac{\sin^2u}{\cos u} \]
Thus:
\[ \frac{du}{dx}=\frac{\sin^2u}{\cos u} \Rightarrow \frac{\cos u}{\sin^2u}\,du=dx \]
Step 3: Integrate both sides.
\[ \int \frac{\cos u}{\sin^2u}\,du=\int dx \]
Let \(w=\sin u\Rightarrow dw=\cos u\,du\).
\[ \int \frac{1}{w^2}\,dw=x+c \]
\[ -\frac{1}{w}=x+c \]
So:
\[ -\frac{1}{\sin u}=x+c \Rightarrow x+\cosec u=c \]
Replace \(u=x+y\):
\[ x+\cosec(x+y)=c \]
Final Answer:
\[ \boxed{x+\cosec(x+y)=c} \]