Question

The general solution of the differential equation $\cos(x + y) \frac{dy}{dx} = 1$ is

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

Hint:

Whenever your substitution integration simplifies to the form $\int \frac{1}{1+\cos v} dv$, it will always generate a half-angle tangent term ($\tan\frac{v}{2}$) due to the identity conversion. Scanning the options for half-angles lets you rule out choices (A) and (B) instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the general solution of a first-order non-separable differential equation where the variables are combined within a cosine function argument.

Step 2: Key Formula or Approach:
Equations of the form $\frac{dy}{dx} = f(ax + by + c)$ can be transformed into a separable form by using a substitution variable:
$$v = x + y \implies \frac{dv}{dx} = 1 + \frac{dy}{dx}$$

Step 3: Detailed Explanation:
Let's substitute $v = x + y \implies \frac{dy}{dx} = \frac{dv}{dx} - 1$ into our differential equation:
$$\cos v \left(\frac{dv}{dx} - 1\right) = 1$$ $$\cos v \frac{dv}{dx} - \cos v = 1 \implies \cos v \frac{dv}{dx} = 1 + \cos v$$ Now, separate the variables $v$ and $x$:
$$\frac{\cos v}{1 + \cos v} \, dv = dx$$ Integrate both sides:
$$\int \frac{\cos v}{1 + \cos v} \, dv = \int dx$$ To simplify the left integral, add and subtract 1 in the numerator:
$$\int \left(\frac{1 + \cos v - 1}{1 + \cos v}\right) dv = \int \left(1 - \frac{1}{1 + \cos v}\right) dv = x + c$$ Using the half-angle trigonometric identity $1 + \cos v = 2 \cos^2\left(\frac{v}{2}\right)$:
$$\int dv - \int \frac{1}{2 \cos^2(v/2)} \, dv = x + c$$ $$v - \frac{1}{2} \int \sec^2\left(\frac{v}{2}\right) dv = x + c$$ We know that $\int \sec^2\left(\frac{v}{2}\right) dv = 2 \tan\left(\frac{v}{2}\right)$:
$$v - \frac{1}{2} \left(2 \tan\left(\frac{v}{2}\right)\right) = x + c \implies v - \tan\left(\frac{v}{2}\right) = x + c$$ Now substitute back $v = x + y$:
$$(x + y) - \tan\left(\frac{x + y}{2}\right) = x + c$$ The $x$ terms on both sides cancel out perfectly:
$$y - \tan\left(\frac{x + y}{2}\right) = c \implies y = \tan\left(\frac{x + y}{2}\right) + c$$ This matches option (C).

Step 4: Final Answer:
The general solution is $y = \tan\left(\frac{x + y}{2}\right) + c$, which corresponds to option (C).