Question

The general solution of differential equation $(y^2 - x^2)dx = xy dy$ ($x \neq 0$) is ______.

• A. $2x^2 \log x + y^2 + 2cx^2 = 0$, where $c$ is the constant of integration
• B. $2x^2 \log x - y^2 + 2cx^2 = 0$, where $c$ is the constant of integration
• C. $x^2 \log x + y^2 + 2cx^2 = 0$, where $c$ is the constant of integration
• D. $x^2 \log x - y^2 + 2cx^2 = 0$, where $c$ is the constant of integration

Hint:

To easily verify if a differential equation is homogeneous, replace $x$ with $kx$ and $y$ with $ky$. If all $k$'s cancel out perfectly, the equation is homogeneous and the substitution $y=vx$ will always work.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have a first-order differential equation. Let's rewrite it in the standard $\frac{dy}{dx}$ format:
$$\frac{dy}{dx} = \frac{y^2 - x^2}{xy}$$
Since every term on the right side has a uniform total degree of 2, this is a homogeneous differential equation.

Step 2: Key Formula or Approach:
For a homogeneous differential equation, we make the substitution $y = vx$.
Differentiating both sides with respect to $x$ using the product rule:
$$\frac{dy}{dx} = v + x\frac{dv}{dx}$$

Step 3: Detailed Explanation:
Substitute $y = vx$ and the derivative into the differential equation:
$$v + x\frac{dv}{dx} = \frac{(vx)^2 - x^2}{x(vx)}$$
Factor out $x^2$ in the numerator and denominator:
$$v + x\frac{dv}{dx} = \frac{x^2(v^2 - 1)}{x^2(v)} = \frac{v^2 - 1}{v}$$
Subtract $v$ from both sides:
$$x\frac{dv}{dx} = \frac{v^2 - 1}{v} - v = \frac{v^2 - 1 - v^2}{v} = -\frac{1}{v}$$
Separate the variables to integrate:
$$v \, dv = -\frac{1}{x} \, dx$$
Integrate both sides:
$$\int v \, dv = -\int \frac{1}{x} \, dx$$
$$\frac{v^2}{2} = -\log|x| + C'$$
Substitute $v = \frac{y}{x}$ back into the equation:
$$\frac{y^2}{2x^2} = -\log x + C'$$
Multiply the entire equation by $2x^2$ to clear the denominator:
$$y^2 = -2x^2 \log x + 2C'x^2$$
Rearrange to group all terms on one side:
$$2x^2 \log x + y^2 - 2C'x^2 = 0$$
Let $c = -C'$ (since $C'$ is just an arbitrary constant):
$$2x^2 \log x + y^2 + 2cx^2 = 0$$

Step 4: Final Answer:
The general solution matches option (a).