Question

The solution of the differential equation \(\frac{dy}{dx} = cos(x+y)\) is...

• A. \(cot(\frac{x+y}{2}) = x+c\)
• B. \(tan(\frac{x+y}{2}) = x+c\)
• C. \(-sin(x+y) = x+c\)
• D. \(sec(x-y)+x = c\)

Hint:

Put v = x + y so that the variables separate.

Answer 1

The Correct Option is B

Step 1: Substitute
Let \(v = x + y\). Then \(\dfrac{dv}{dx} = 1 + \dfrac{dy}{dx} = 1 + \cos v\).

Step 2: Use a half angle identity
\(1 + \cos v = 2\cos^2\dfrac{v}{2}\). So \(\dfrac{dv}{dx} = 2\cos^2\dfrac{v}{2}\).

Step 3: Separate and integrate
\[ \int\frac{dv}{2\cos^2(v/2)} = \int dx \Rightarrow \int\frac{1}{2}\sec^2\frac{v}{2}\,dv = x + c \]
\[ \tan\frac{v}{2} = x + c \]

Step 4: Result
Resubstitute \(v = x + y\): \(\tan\dfrac{x+y}{2} = x + c\), option (B). Option (A) uses cot, whose derivative has the wrong sign, and the others do not reduce to this form.

Final Answer:
The solution is tan((x + y)/2) = x + c. This is option (B). \[ \boxed{\text{(B) }\tan\frac{x+y}{2}=x+c} \]