Q.20. The differential equation \cos(x+y)\, dy = dx has the general solution:
A. y=\sin(x+y)+C, where C is a constant.
B. y=\tan(x+y)+C, where C is a constant.
C. y=\tan\left(\frac{x+y}{2}\right)+C, where C is a constant.
D. y=\frac{1}{2}tan (x+y)+c
Answer :- C. y=\tan\left(\frac{x+y}{2}\right)+C, where C is a constant.
Explanation :- cos(x+y) dy = dx
\frac{dx}{dy}= cos (x+y) ——(i)
Put x+y u —-(i)
Differentiating w.r.t y we get
\frac{dx}{dy}+1= \frac{du}{dy}\frac{dx}{dy}= \frac{du}{dy}-1 –(iii)
Substituting (ii) and (iii) in (i) we get
\frac{du}{dy}-1 = cos\; u \frac{du}{1+ cos\; u} = dy \frac{du}{2 cos^{2}\left ( \frac{u}{2} \right )} = dyIntegrating on both sides, we get
\frac{1}{2}\int sec^{2}\left ( \frac{u}{2} \right )du=\int dy y=tan\left ( \frac{x+y}{2} \right )+c