• # question_answer Let y= y(x) be the solution curve of the differential equation, $({{y}^{2}}-x)\frac{dx}{dx}=1,$ satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is [JEE MAIN Held on 07-01-2020 Evening] A) 2 - e B) 2 + e C) - e                    D) 2

[a] $\because \,\,\,({{y}^{2}}-x)\frac{dy}{dx}=1$ $\therefore \,\,\,\,\,{{y}^{2}}dy-xdy=dx$ $\Rightarrow \,\,{{e}^{y}}\cdot {{y}^{2}}dy={{e}^{y}}dx+x{{e}^{y}}dy$ $\Rightarrow \,\,{{e}^{y}}\cdot {{y}^{2}}dy=d({{e}^{y}}\cdot x)$ On integrating both sides we get $\int{{{e}^{y}}\cdot {{y}^{2}}dy=\int{d({{e}^{y}}\cdot x)}}$ ${{y}^{2}}\cdot {{e}^{y}}-\int{2y\cdot {{e}^{y}}dy}={{e}^{y}}\cdot x$ ${{y}^{2}}\cdot {{e}^{y}}-2\left\{ y\cdot {{e}^{y}}-\int{{{e}^{y}}dy} \right\}={{e}^{y}}\cdot x$ $\therefore \,\,\,{{y}^{2}}\cdot {{e}^{y}}-2y{{e}^{y}}+2{{e}^{y}}+2{{e}^{y}}={{e}^{y}}\cdot x+c$ $\therefore \,\,\,\,\,y(0)=1$ $\Rightarrow \,\,c=e$ $\therefore \,\,{{y}^{2}}-2y+2=x+e\cdot {{e}^{-y}}$ $\therefore$    when y = 0 then x = 2 - e