Question

The solution of \(\frac{\text{d}y}{\text{d}x} = sin(x+y)+cos(x+y)\) is

• A. \(log[1+tan(\frac{x+y}{2})] = y+c\), where c is the constant of integration
• B. \(log[1-tan(\frac{x+y}{2})] = y+c\), where c is the constant of integration
• C. \(log[1+tan(\frac{x+y}{2})] = x+c\), where c is the constant of integration
• D. \(log[1-tan(\frac{x+y}{2})] = x+c\), where c is the constant of integration

Hint:

Substitute v = x + y and use the tangent half angle substitution.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Put \(v = x + y\). Then \(\dfrac{dv}{dx} = 1 + \dfrac{dy}{dx} = 1 + \sin v + \cos v\), which is a variable separable form.

Step 2: Half angle substitution
Let \(t = \tan\frac v2\), so \(\sin v = \dfrac{2t}{1 + t^2}\), \(\cos v = \dfrac{1 - t^2}{1 + t^2}\).
\[ 1 + \sin v + \cos v = \frac{(1 + t^2) + 2t + (1 - t^2)}{1 + t^2} = \frac{2(1 + t)}{1 + t^2} \]
Also \(dv = \dfrac{2\,dt}{1 + t^2}\).

Step 3: Integrate
\[ dx = \frac{dv}{1 + \sin v + \cos v} = \frac{2\,dt/(1 + t^2)}{2(1 + t)/(1 + t^2)} = \frac{dt}{1 + t} \]
\[ x + c = \log(1 + t) \Rightarrow \log\left[1 + \tan\left(\frac{x + y}{2}\right)\right] = x + c \]
This is option (C). Option (A) has y on the right side, which would not follow since the left side was divided by dx.

Final Answer:
The solution is \(\log\left[1 + \tan\frac{x + y}{2}\right] = x + c\), option (C). \[ \boxed{\log\left[1 + \tan\left(\frac{x+y}{2}\right)\right] = x + c} \]