Respuesta :
Answer: Theorem 1.2.1 DOES guarantee that this differential equation has a unique solution through the given point.
Step-by-step explanation: The equation is [tex]y'=\sqrt{y^2-25}[/tex]. We identify [tex]f(x,y)=\sqrt{y^2-25}[/tex]. The theorem 1.2.1. asks us to find the rectangular region R such that [tex]f(x,y)[/tex] and [tex]\frac{\partial f}{\partial y}[/tex] are both continuous on that region. We find that
[tex]\frac{\partial f}{\partial y}=\frac{y}{\sqrt{y^2-25}}.[/tex]
The continuity of both these functions depens on the factor [tex]\sqrt{y^2-25}[/tex]. It is real valued for [tex]y\leq 5[/tex] and [tex]y\geq 5[/tex]. Our point has [tex]y=6[/tex] so we take the region where [tex]y\geq5[/tex]. In the partial derivative this factor is in the denominator so to ensure that it is defined we exclude the possibility of y=5 i.e. we take any interval [tex]y\in[a,b][/tex], where [\tex]a>5[tex] and [tex]b>6[/tex] ([tex]b>a[/tex] of course). [tex]f(x,y)[/tex] is independent of [tex]x[/tex] so both it and its partial derivative wrt y are continuous on any interval for x. This means that we can take any interval [tex]x\in[c,d][/tex] that contains x=1. So on the rectangle [tex]a\leq y\leq b[/tex] and [tex]c<x<d[/tex] such that [tex]a>5, b>a,6, c<1,d>1[/tex] (the rectangle includes the given point [tex](x_0,y_0)=(1,6)[/tex]), both [tex]f(x,y)[/tex] and [tex]\frac{\partial f}{\partial y}[/tex] are continuous, and, therefore, there is an interval [tex]x\in[1-h,1+h][/tex] such that there is a unique sulution to the given initial value problem, according to the theorem 1.2.1.
