In the first group of problems determine a set \(S\) satisfying the hypotheses of the Existence and Uniqueness Theorem posed in this section.
\(y' = \frac{1}{1-t^2}; \,\, y(0) = 1\)
\(\dot{y} = \frac{1}{y-t^2}; \,\, y(0) = 1\)
\(y' = \sqrt{yt}; \,\, y(1) = 1\)
\(y' = \log(1-t^2-y^2); \,\, y(0) = 0\)
\(t\dot{y} = \sqrt{e^y - 1}; \,\, y(1) = 1\)
\(yy' = \log(t); \, \, y(1) = 1\)
\(y' = \frac{y+t}{y^2-1}; \,\, y(0) = 0\)
\(\frac{1}{t^2} + \frac{w'}{\sqrt{w}}=0; \, \, w(-1)=1.\)
\(\dot{x} = \frac{\tan(t)}{x^2+2tx+t^2}; \,\, x(0)=1\)
\(\theta' + \log(t) \theta = \sqrt{4-t^2}; \, \, \theta(1) = 2\)
\(\dot{y} = \frac{\sqrt{x^2-y^2}}{\log(y)}; \,\, y(-1) = 1/2\)
\(x' = \frac{(\log(t))^2}{x^2 - 2x -3}; \,\, x(1) =0\)
For the next few problems be sure to justify your responses.
A linear first order equation in normal form is given by \(y' + p(t)y = q(t)\) with initial condition \(y(t_I) = y_I\). Use the general existence and uniqueness theorem to justify why we can guarantee a unique solution when both \(p\) and \(q\) are continuous function on an open interval containing \(t_I\).
Suppose we have a separable equation \(y' = \frac{g(t)}{h(y)}\) with initial condition \(y(t_I) = y_I\). What conditions must we place on \(g\), \(h\), \(t_I\) and \(y_I\) for the solution to be unique?