(4−y2)y′=x2
According to theorem of Existence of Unique Solution if f(x,y) and df/dy are continious
on rectangular region R theh there is exist interval I on which unique exists.
We have:
f(x,y)=4−y2x2
dydf=(4−y2)22yx2
f(x,y) and df/dy are not continious at y=−2 and y=2
Answer: C
A unique solution exists in the region consisting of all points in the xy-plane except (0,2) and (0,−2)
Comments