Question

The general solution of the differential equation \( \left(x - (x + y)\log(x + y)\right) dx + x\,dy = 0 \) is:

• A. \( y \log(x + y) = cx \)
• B. \( x \log(x + y) = cy \)
• C. \( \log(x + y) = cy \)
• D. \( \log(x + y) = cx \)

Hint:

Look for substitution opportunities (like \(x + y\)) when complex expressions appear repeatedly. That often simplifies the differential equation significantly.

Answer 1

The Correct Option is D

Step 1: Write the given differential equation. \[ \left(x - (x + y)\log(x + y)\right) dx + x\,dy = 0 \] Step 2: Simplify the first term. \[ x - (x + y)\log(x + y) = x - x\log(x + y) - y\log(x + y) = x(1 - \log(x + y)) - y\log(x + y) \] So the equation becomes: \[ [x(1 - \log(x + y)) - y\log(x + y)] dx + x\,dy = 0 \tag{1} \] Step 3: Use substitution. Let \[ u = x + y \quad \Rightarrow \quad du = dx + dy \Rightarrow dy = du - dx \] \[ Also, \quad y = u - x \] Now plug into the differential equation: \[ \left[x - u \log u\right] dx + x(dy) = 0 \Rightarrow \left[x - u \log u\right] dx + x(du - dx) = 0 \] Step 4: Simplify the equation. \[ [x - u \log u] dx + x\,du - x\,dx = 0 \Rightarrow -u \log u \, dx + x\,du = 0 \] Step 5: Rearranging gives \[ x\,du = u \log u\, dx \Rightarrow \frac{du}{\log u} = \frac{u}{x} dx \] Step 6: Separate and integrate. Bring all \(u\) terms to one side: \[ \frac{1}{u \log u} du = \frac{1}{x} dx \] Integrate both sides: \[ \int \frac{1}{u \log u} du = \int \frac{1}{x} dx \Rightarrow \log(\log u) = \log x + \log C = \log(Cx) \] \[ \Rightarrow \log u = Cx \quad (\text{exponentiate both sides}) \Rightarrow \log(x + y) = Cx \]