Chapter 2.4, Problem 14P

Verify that both $y_1\left(t\right)=1-t$ and $y_2\left(t\right)=-t^2/4$ are solutions of the initial vale problem

$y\text{'}=\frac{-t+{\left(t^2+4y\right)}^{1/2}}{2},y\left(2\right)=-1$

Where are these solutions valid?

Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of Theorem $\mathrm{2.4.2}$.

Show that $y=ct+c^2$, where $c$ is an arbitrary constant, satisfies the differential equation in part (a) for $t\ge -2c$. If $c=-1$, the initial condition is also satisfied, and the solution $y=y_1\left(t\right)$ is obtained. Show that there is no choice of $c$ that gives the second solution $y=y_2\left(t\right)$.

Theorem $\mathrm{2.4.2}$:

Let the functions $f$ and $\partial f/\partial y$ be continuous in some rectangle $\alpha <t<\beta ,\gamma <t<\delta$ containing the point $\left(t_0,y_0\right)$. Then, in some interval $t_0-h<t<t_0+h$ contained in $\alpha <t<\beta$, there is unique solution $y=\varphi \left(t\right)$ of the initial value problem

$y\text{'}=f\left(t,y\right),y\left(t_0\right)=y_0$.