Question

The general solution of the differential equation $x \, dy - y \, dx = 0$ is:

• A. $x^2 - y^2 = k$
• B. $xy = k$
• C. $x = ky$
• D. $\log y + \log x = k$

Hint:

You can also identify this using the quotient rule structure! Remember that $d\left(\frac{y}{x}\right) = \frac{x\,dy - y\,dx}{x^2}$. Since the numerator is 0, $d\left(\frac{y}{x}\right) = 0 \Rightarrow \frac{y}{x} = c \Rightarrow y = cx$, which matches $x = ky$.

Answer 1

The Correct Option is C

Concept: This is a first-order, first-degree differential equation that can be solved easily using the variable separable method. In this method, all terms involving $y$ are moved with $dy$ on one side, and all terms involving $x$ are moved with $dx$ on the other side before integrating.

Step 1: Separate the variables $x$ and $y$.
The given differential equation is: \[ x \, dy - y \, dx = 0 \] Rearranging terms by moving $y \, dx$ to the right-hand side: \[ x \, dy = y \, dx \] Assuming $x \neq 0$ and $y \neq 0$, divide both sides by $xy$ to separate the variables: \[ \frac{dy}{y} = \frac{dx}{x} \]

Step 2: Integrate both sides of the equation.
Now, integrate both sides with respect to their corresponding variables: \[ \int \frac{1}{y} \, dy = \int \frac{1}{x} \, dx \] Using the standard rule $\int \frac{1}{u} \, du = \log|u|$, we get: \[ \log|y| = \log|x| + C \] where $C$ is an arbitrary constant of integration.

Step 3: Simplify the expression using logarithmic properties.
Let us express the constant $C$ as $\log|c|$ (where $c$ is a positive constant) to group it easily with the other logarithmic terms: \[ \log|y| = \log|x| + \log|c| \] Using the log property $\log(A) + \log(B) = \log(AB)$, we can combine the right side: \[ \log|y| = \log|c \cdot x| \] Taking the exponential on both sides (removing logs): \[ y = cx \quad \Rightarrow \quad x = \frac{1}{c}y \] Let $\frac{1}{c} = k$ be a new arbitrary constant. Then: \[ x = ky \] This represents a family of straight lines passing through the origin.