Question

Determine whether Theorem 1.2.1 guarantees that the differential equation y' = y2 â 25 possesses a unique solution through the given point. (1, 6)

Answer 1

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
`y'=√(y^2-25)`. We identify
`f(x,y)=√(y^2-25)`. The theorem 1.2.1. asks us to find the rectangular region R such that
`f(x,y)` and
`(\partial f)/(\partial y)` are both continuous on that region. We find that

`(\partial f)/(\partial y)=(y)/(√(y^2-25)).`

The continuity of both these functions depens on the factor
`√(y^2-25)`. It is real valued for
`y\leq 5` and
`y\geq 5`. Our point has
`y=6` so we take the region where
`y\geq5`. 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
`y\in[a,b]`, where [\tex]a>5
`and [tex]b>6` (
`b>a` of course).
`f(x,y)` is independent of
`x` so both it and its partial derivative wrt y are continuous on any interval for x. This means that we can take any interval
`x\in[c,d]` that contains x=1. So on the rectangle
`a\leq y\leq b` and
`c<x<d` such that
`a>5, b>a,6, c<1,d>1` (the rectangle includes the given point
`(x_0,y_0)=(1,6)`), both
`f(x,y)` and
`(\partial f)/(\partial y)` are continuous, and, therefore, there is an interval
`x\in[1-h,1+h]` such that there is a unique sulution to the given initial value problem, according to the theorem 1.2.1.