Here is a problem, this is in my compulsory list of homework, I'm trying to solve but it seems very hard for me, I've tried Taylor expansion, considered some cases of the value c,... Can you give me some hints to solve ?
- Consider the problem
$Y' = \dfrac{1}{t+1}+c\cdot\tan^{-1}(Y(t))-\dfrac{1}{2},\quad Y(0)=0$
with $c$ a given constant. Since $Y'(0) = \dfrac{1}{2}$, the solution Y (t) is initially increasing as $t$ increases, regardless of the value of $c$. As best you can, show that there is a value of $c$, call it $c^∗$, for which (1) if $c > c^∗$, the solution $Y(t)$ increases indefinitely, and (2) if $c < c^∗$, then $Y(t)$ increases initially, but then peaks and decreases. Using ode45, determine $c^∗$ to within 0.00005, and then calculate the associated solution $Y(t)$ for $0 \leq t \leq 50$.
It's in problems of chapter 5 : Taylor and Runge–Kutta methods of NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS, Kendall Atkinson, Weimin Han, David Stewart, University of Iowa, Iowa City, Iowa.
Here is the link of this book : https://homepage.math.uiowa.edu/~atkinson/papers/NAODE_Book.pdf
Many thanks.
$c=0$ can be integrated directly, $$y(t)=y_0(t)=\ln(1+t)-\frac t2,$$ and the value at $t=e^2-1>6$ is negative.
For any $c>0$, the observed initial slope gives an initial segment where $y$ is positive.
For $c\approx\infty$, even a small positive value of $y$ will lead to a large slope catapulting the $y$ curve towards very large values. After that the other two terms do not matter any more, the slope remains positive, $y$ does not come down, back from "almost infinity".
One could try to make the last point a bit more quantitative by replacing "almost infinity" with some very large value, e.g., $c=1000$, and estimate the magnitude of all terms for $t=0.01$, to then find the announced behavior.