Q100 / 150 · MHT CET 2025 (25 Apr Shift 2) · Mathematics
The general solution of differential equation $\left(y^2-x^2\right) \mathrm{d} x=x y$ dy $(x \neq 0)$ is
- Correct answerA$2 x^2 \log x+\mathrm{y}^2+2 \mathrm{c} x^2=0$, where c is the constant of integration
- B$2 x^2 \log x-\mathrm{y}^2+2 \mathrm{c} x^2=0$, where c is the constant of integration
- C$x^2 \log x+\mathrm{y}^2+2 \mathrm{c} x^2=0$, where c is the constant of integration
- D$x^2 \log x-\mathrm{y}^2+2 \mathrm{c} x^2=0$, where c is the constant of integration
Solution
Given the homogeneous differential equation $\left(y^2-x^2\right) dx = x y dy$, express $\frac{dy}{dx}$ as $\frac{y^2-x^2}{xy}$.
The substitution $y = vx$ with $\frac{dy}{dx} = v + x\frac{dv}{dx}$ simplifies to $v + x\frac{dv}{dx} = \frac{v^2-1}{v}$.
Rearranging gives $x\frac{dv}{dx} = \frac{-1}{v}$, so $v dv = -\frac{1}{x} dx$.
Integration yields $\frac{v^2}{2} = -\ln|x| + C'$.
Substituting back $v = \frac{y}{x}$, we obtain $\frac{y^2}{2x^2} = -\ln|x| + C'$.
Multiplying through by $2x^2$ and rearranging, $y^2 + 2x^2 \ln x - 2C'x^2 = 0$.
Letting $2C' = -2c$ gives $y^2 + 2x^2 \ln x + 2cx^2 = 0$, which corresponds to option A.
Final answer: A
Hence, the correct answer is option: 'A'.