Q. The solution of the differential equation $\frac{d y}{d x}+\frac{1}{x}=\frac{e^{y}}{x^{2}}, \, \left(\forall x > 0\right)$ is $\lambda xe^{- y}=1-2cx^{2}$ (where $c$ is an arbitrary constant). Then, the value of $\lambda $ is equal to
NTA AbhyasNTA Abhyas 2022
Solution: