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

$\frac{\partial f}{\partial y}=\frac{y}{\sqrt{y^2-25}}.$

The continuity of both these functions depens on the factor $\sqrt{y^2-25}$. 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 $c$ such that $a>5, b>a,6, c1$ (the rectangle includes the given point $(x_0,y_0)=(1,6)$), both $f(x,y)$ and $\frac{\partial f}{\partial y}$ 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.