MHTCET 25 Apr 2025 Shift 2 Paper

Concept:This is a homogeneous differential equation. Use the substitution $y = vx$ to make it separable.Explanation:Given:$$\left(y^{2} - x^{2}\right) dx = xy \, dy$$Since $x \neq 0$, rewrite as:$$\frac{dy}{dx} = \frac{y^{2} - x^{2}}{xy}$$Put $y = vx$.Then:$$\frac{dy}{dx} = v + x \frac{dv}{dx}$$Substitute into the differential equation:$$v + x \frac{dv}{dx} = \frac{v^{2}x^{2} - x^{2}}{x(vx)} = \frac{v^{2} - 1}{v}$$So:$$x \frac{dv}{dx} = \frac{v^{2} - 1}{v} - v = -\frac{1}{v}$$Separate the variables:$$v \, dv = -\frac{dx}{x}$$Integrate both sides:$$\frac{v^{2}}{2} = -\ln|x| + c$$Replace $v = \frac{y}{x}$:$$\frac{y^{2}}{2x^{2}} = -\ln|x| + c$$Multiply by $2x^{2}$:$$y^{2} = -2x^{2}\ln|x| + 2cx^{2}$$Rearranging:$$2x^{2}\log x + y^{2} + 2kx^{2} = 0$$where $k = -c$ is an arbitrary constant.This matches the form given in option A.Answer:Option A: $2x^{2}\log x + y^{2} + 2cx^{2} = 0$, where $c$ is the constant of integration.