Question

The general solution of the differential equation \( \cos(x + y) \, dy = dx \) is

• A. \( y = \tan \left( \frac{x + y}{2} \right) + c \)
• B. \( y = \sec \left( \frac{x + y}{2} \right) + c \)
• C. \( y = x \sec \left( \frac{y}{x} \right) + c \)
• D. \( y = -\cos^{-1} \left( \frac{y}{x} \right) + c \)

Hint:

When dealing with implicit functions like \( \cos(x + y) \, dy = dx \), consider using substitution \( u = x + y \) to simplify the equation.

Answer 1

The Correct Option is A

We are given the differential equation: \[ \cos(x + y) \, dy = dx \] Step 1: Separate variables. \[ \cos(x + y) \, dy = dx
\Rightarrow
\cos(x + y) = \frac{dx}{dy} \] Let us try substitution: \( u = x + y \Rightarrow \frac{du}{dy} = \frac{dx}{dy} + 1 \) So, \[ \cos(u) = \frac{dx}{dy} = \frac{du}{dy} - 1 \Rightarrow \cos(u) = \frac{du}{dy} - 1 \Rightarrow \frac{du}{dy} = \cos(u) + 1 \] Step 2: Separate and integrate. \[ \frac{du}{\cos(u) + 1} = dy \] Use the identity: \[ \frac{1}{\cos u + 1} = \frac{\sec^2\left( \frac{u}{2} \right)}{2} \] So, \[ \int \frac{du}{\cos u + 1} = \int \frac{\sec^2\left( \frac{u}{2} \right)}{2} \, du = \int \sec^2\left( \frac{u}{2} \right) . \frac{1}{2} \, du \] Let \( w = \frac{u}{2} \Rightarrow du = 2 \, dw \) \[ \int \sec^2(w) \, dw = \tan(w) + C = \tan\left( \frac{u}{2} \right) + C \] So, \[ y = \tan\left( \frac{x + y}{2} \right) + c \] % Final Answer \[ \boxed{ y = \tan\left( \frac{x + y}{2} \right) + c } \]