We have the equation dxdy=−yx with initial condition y(2)=−π. One of solutions for this equation is x2+y2=π2+4 . It can be transformes into two functions: y(x)=π2+4−x2 and y(x)=−π2+4−x2 .Function y(x)=π2+4−x2 does not satisfy the initial condition y(2)=−π . The second function,
y(x)=−π2+4−x2 , defined on (−π2+4,π2+4), satisfies the initial condition.
From https://www.science.unitn.it/~bagagiol/noteODE.pdf, theorem 2.10 we have that for all a∈(0,π) the equation has a unique local solution in (−π2+4,π2+4)×(−∞,a) (which is y=−π2+4−x2 ).
Hence the equation has a unique local solution on (−π2+4,π2+4)×(−∞,0) .Indeed, assume that there is solution y1(x) defined on I , where y1(u)=−π2+4−u2 for some u∈(−π2+4,π2+4)∩I . Ifv=max{y1(u),−π2+4−u2} , we have two different solutions in (I∩(−π2+4,π2+4))×(−∞,2v) . Contradiction with uniqueness of y(x)=−π2+4−x2.
Assume that there is solution y2(x), defined on I1 such that I1∖[−π2+4,π2+4]=∅ . If there is c∈I1∩(−∞,−π2+4) , then [c,2]⊂I1 and y2(x)=−π2+4−x2 on [−π2+4,2] . So y2(−π2+4)=0 and ∣∣y2′(−π2+4)∣∣=+∞.
Let x∈(c,−π2+4). Since y2′(x)=−y2(x)x and x<0 , we have sgn(y2′(x))=sgn(y2(x)) .
If y2′(−π2+4−0)=−∞ , then y′(x)<0 in (d,−π2+4) for some d∈(c,−π2+4) . But since y2 is decreasing function in (d,−π2+4) and y2(−π2+4)=0 , we have y2>0 in (d,−π2+4) . Contradiction with statement that sgn(y2′)=sgn(y2) in (d,−π2+4) . Similarly we obtain a contradiction while assuming that y2′(−π2+4−0)=+∞. So I1∩(−∞,−π2+4)=∅ .
Likewise we can obtain that I1∩(π2+4,+∞)=∅ .
Hence every solution of the equation is defined on a subset of [−π2+4,π2+4] . Uniqueness of the local solution of y(x)=−π2+4−x2 in [−π2+4,π2+4]×(−∞,0] gives us uniqueness of the solution y(x)=−π2+4−x2 in R2 .
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