Question 224 of 490MediumMCQ
Let y=y(x) solve dxdy=1+xey−x, −2<x<2, y(0)=0. The minimum value of y(x) on (−2,2) is: Solutions · 1
Official solution
\[\frac{dy-dx}{e^{y-x}} = x\,dx \implies -e^{x-y} = \frac{x^2}{2}+c.\]
Using (0,0): c=−1. So ex−y=22−x2, i.e. y=x−ln22−x2.
\[\frac{dy}{dx} = 1+\frac{2x}{2-x^2} = \frac{2+2x-x^2}{2-x^2}=0 \implies x=1-\sqrt{3}.\]
\[y(1-\sqrt{3}) = (1-\sqrt{3})-\ln(\sqrt{3}-1).\] Answer the question to unlock