Question

The general solution of the differential equation $x \, dy - y \, dx = y^2 \, dx$ is:

• A. $y = \frac{x}{C - x}$
• B. $x = \frac{2y}{C + x}$
• C. $y = (C + x)(2x)$
• D. $y = \frac{2x}{C + x}$
• E. $x = \frac{y}{C - x}$

Hint:

When you see $x \, dy - y \, dx$, try dividing by $y^2$ to get $-d(x/y)$ or by $x^2$ to get $d(y/x)$. Here, dividing by $y^2$ gives $d(x/y) = -dx/x$ if rearranged slightly, which is even faster.

Answer 1

The Correct Option is A

Concept: This equation can be solved by rearranging it into a form where variables can be separated or by recognizing a common differential.

Step 1: Rearrange to separate variables.
\[ x \, dy = (y^2 + y) \, dx \] Divide by $x(y^2 + y)$: \[ \frac{dy}{y^2 + y} = \frac{dx}{x} \]

Step 2: Integrate both sides.
Use partial fractions for the L.H.S: $\frac{1}{y(y+1)} = \frac{1}{y} - \frac{1}{y+1}$. \[ \int \left( \frac{1}{y} - \frac{1}{y+1} \right) dy = \int \frac{1}{x} dx \] \[ \log|y| - \log|y+1| = \log|x| + \log|C| \] \[ \log \left| \frac{y}{y+1} \right| = \log|Cx| \] \[ \frac{y}{y+1} = Cx \]

Step 3: Solve for $y$.
\[ y = Cx(y + 1) = Cxy + Cx \] \[ y(1 - Cx) = Cx \] \[ y = \frac{Cx}{1 - Cx} \] Divide numerator and denominator by $C$ (let $1/C = K$): \[ y = \frac{x}{K - x} \] Replacing $K$ with $C$ for the general constant gives: \[ y = \frac{x}{C - x} \]