MHTCET 23 Apr 2025 Shift 1 Paper

Concept:Use the substitution $v=\frac{y}{x}$ to convert the given homogeneous differential equation into a separable form.Explanation:Given equation is:$$x \frac{dy}{dx}=y(\log y-\log x+1)$$Since $\log y-\log x=\log\left(\frac{y}{x}\right)$, we get:$$\frac{dy}{dx}=\frac{y}{x}\left[\log\left(\frac{y}{x}\right)+1\right]$$Put $v=\frac{y}{x}$, so $y=vx$.Differentiate: $\frac{dy}{dx}=v+x\frac{dv}{dx}$.Substitute into the equation:$$v+x\frac{dv}{dx}=v(\log v+1)$$Simplify:$$x\frac{dv}{dx}=v\log v$$Separate the variables:$$\frac{dv}{v\log v}=\frac{dx}{x}$$Integrating both sides:$$\int \frac{dv}{v\log v}=\int \frac{dx}{x}$$Let $u=\log v$, so $du=\frac{dv}{v}$.Then $\int \frac{du}{u}=\log x+\log c$.Thus $\log(\log v)=\log(cx)$.Hence $\log v=cx$.Replacing $v=\frac{y}{x}$:$$\log\left(\frac{y}{x}\right)=cx$$This matches option D.Answer:Option D: $\log \frac{y}{x}=cx$, where $c$ is the constant of integration.