COMEDK 2025 Morning Solved Paper

Concept:This is an exact differential equation. We find a potential function $\Phi(x,y)$ and apply the initial condition to determine the constant.Explanation:The given equation is $y\,dx + (x+\ln y)\,dy = 0$.Check exactness: $\frac{\partial}{\partial y}(y)=1$ and $\frac{\partial}{\partial x}(x+\ln y)=1$, so it is exact.Let $\Phi(x,y)$ satisfy $\frac{\partial \Phi}{\partial x}=y$. Then $\Phi=xy + h(y)$.Differentiate w.r.t. $y$: $\frac{\partial \Phi}{\partial y}=x + h'(y)$. This must equal $x+\ln y$.Thus $h'(y)=\ln y$, so $h(y)=y\ln y - y$ (since $\int \ln y\,dy = y\ln y - y$).Hence the general solution is $\Phi=xy + y\ln y - y = C$.Use $y(0)=1$: $0\cdot1 + 1\cdot\ln 1 - 1 = C \Rightarrow C=-1$.Therefore $xy + y\ln y - y = -1$, which rearranges to $y(x-1+\ln y)+1=0$.Answer:Option A: $y(x-1+\log y)+1=0$.