Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations3 Marks
Question
Solve : $\cos (x+y) d y=d x$.
✓
Answer
Writing the given equation in the following form:
$\frac{d y}{d x}=\frac{1}{\cos (x+y)}$
Here it is clear that in equation $1 ,$ we cannot separate the variables $x$ and $y$ until we substitute $x+y=t$.
Hence taking $x+y=t$
$\therefore 1+\frac{d y}{d x}=\frac{d t}{d x} \therefore \frac{d y}{d x}=\frac{d t}{d x}-1$
Hence from equation $(1),$
$\frac{d t}{d x}-1=\frac{1}{\cos t}$
$\Rightarrow \frac{d t}{d x}=\frac{1}{\cos t}+1=\frac{1+\cos t}{\cos t}$
Now we shall separate the variables
$\frac{\cos t}{1+\cos t} d t=d x$
Integrating both sides of equation $(2),$
$\int d x=\int \frac{\cos t}{1+\cos t} d t$
$\int d x=\int\left(1-\frac{1}{1+\cos t}\right) d t$
${rlrl} \Rightarrow \int d x =\int d t-\int \frac{1}{2 \cos ^2 t / 2} d t$
$\Rightarrow \int d x =\int d t-\frac{1}{2} \int \sec ^2 t / 2 d t$
$\Rightarrow x =t-\tan t / 2+C_1$
Putting the value of $t$
$x =x+y-\tan \left(\frac{x+y}{2}\right)+C_1$
$\Rightarrow y =\tan \left(\frac{x+y}{2}\right)-C_1$
Hence $y=\tan \left(\frac{x+y}{2}\right)+C \text { where } C=-C_1$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.