Section 2.4 – Differences Between Linear and Nonlinear Equations
Exercises
Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.
1. Determine an interval in which the solution of the following initial value problem is certain to exist.
\[(t^2-1)y'+(\sin t)y=\frac{\cot t}{t^2-4t+3},\qquad y(2)=-1\]
\(I=(1,3)\)
2. State where in the \(ty\)-plane the hypothesis of the Existence and Uniqueness theorem are satisfied for the following differential equations.
- \(y'=\dfrac{\ln(ty)}{1-(t^2+y^2)}\)
- \(y'=(t^2-y)^{1/3}\)
- Anywhere in the first or third quadrant, but not on the unit circle.
- Any point off the parabola \(y=t^2.\)
3. Solve the following initial value problems and determine how the interval in which the solution exists depends on the initial value \(y_0\).
- \(\displaystyle y'=\frac{-4}{t}y,\qquad\) \(y(2)=y_0\)
- \(\displaystyle y'+y^3=0\qquad\) \(y(t_0)=y_0\)
- The solution is \(y=\dfrac{16y_0}{t^4}\) and \(I.V.=(0,\infty)\). Note that if \(y_0=0\), then \(y=0\) is an equilibrium solution.
- If \(y_0=0\), then \(y=0\) is an equilibrium solution with \(I=(-\infty,\infty)\). If \(y_0>0\), then \(y=\dfrac{1}{\sqrt{2(t-t_0)+\frac{1}{y_0^2}}}\) with \(I=\left(t_0-\dfrac{1}{2{y_0}^2},\infty\right).\) If \(y_0<0\), then \(y=-\dfrac{1}{\sqrt{2(t-t_0)+\frac{1}{y_0^2}}}\) with \(I=\left(t_0-\dfrac{1}{2{y_0}^2},\infty\right).\)
4. Verify that both \(y_1=1-t\) and \(\displaystyle y_2=\frac{-t^2}{4}\) are solutions to the same initial value problem \[y'(t)=\frac{-t+\sqrt{t^2+4y}}{2},\qquad y(2)=-1.\]Does it contradict the existence and uniqueness theorem?
Video Errata: There is a mistake at the 11:00 mark when the presenter wrote out \(f_y\), we should have that \(f_y=\dfrac{1}{2}\cdot\dfrac{1}{2}(t^2+4y)^{-\frac{1}{2}}\cdot (4)=\dfrac{1}{\sqrt{t^2+4y}}.\) This does not change our answer as the condition on \(f_y\) is still \(t^2+4y>0.\)
There is no contradiction because the conditions of the existence and uniqueness theorem are only satisfied for points above the parabola \(y=-\dfrac{t^2}{4}\), and the point \((2,1)\) is on the parabola. For verification that \(y_1\) and \(y_2\) are solutions to the same initial value problem, see video below.