For problems #1–#5a, describe the behavior of the solutions of the autonomous equations by drawing a phase-line portrait. Be sure to comment on the nature of the stationary points.
\(\displaystyle\frac{\dy}{\dt} = y^2 - 1\)
\(\displaystyle \dot{y} = 3 y^3 + 6 y^2\)
\(\displaystyle \frac{\dy}{\dt} = -y^2(1-y)(3-y)\)
\(\displaystyle \frac{\dz}{\dt} = (z-3)^2(z+3)^2\)
\(\displaystyle \frac{\dy}{\dt} = e^y\)
The equation in part (a) is separable; solve it. Comment on whether the explicit solution agrees with your phase line.
Suppose \(f(t)\) is the solution to \[\frac{\dy}{\dt} = (y - 2)(y - 1)^2 (y + 1) (y + 3)^3\] with \(f(0) = -2\). What is \(\lim_{t \to \infty} f(t)\)?
Same differential equation, but change the initial condition to \(f(5) = 0\). What is \(\lim_{t \to -\infty} f(t)\)?
Change again to \(f(-3) = 2\). What is \(\lim_{t \to \infty} f(t)\)?
For problems #7–#9, solve the given initial value problem, then plot the solution you get over the interval given.
\(\dot{w} + w = 2e^t\), \(w(0) = 3\), \(t \in [-5, 5]\)
\(x' - x = t^2\), \(x(0) = 1\), \(t \in [-4, 4]\)
\(y' - 2y = 5\sin(x)\), \(y(0) = 0\), \(t \in [-5, 2]\)
For problems #10–#12, find the general solution to the differential equation, then plot the solutions for the suggested values of \(q\). Do the various solutions all have the same end behavior (i.e., their limits as \(t \to \infty\))?
\(y' + y = 2e^t\), \(y(0) = q\), \(q = -2, -1, 0, 1\), \(t \in [-5, 5]\)
\(\dot{y} - y = t^2\), \(y(0) = q\),\(q = -6, -2, 2, 6\), \(t \in [-4, 4]\)
\(y' - 2y = 5\sin(x)\), \(y(0) = q\),\(q = -3, -1, 1, 3\), \(t \in [-5, 2]\)
Produce a contour plot of \(H(x, y) = \sin(xy)\) on \([-2, 2] \times [-2, 2]\).
Produce a contour plot of \(H(x, y) = x^2 + y^3 + 2xy + 1\) on \([-1.5, 1.5) \times [-1.5, 1.5]\).
Consider the initial value problem \[\frac{\dx}{\dt} = \frac{2 t + 2 t e^x}{e^x} \,, \quad x(1) = 0 \,.\]
Separate the equation and show that the solution to this differential equation satisfies \(\ln(1+e^x) = t^2 + \ln(2) - 1\).
Use a contour plot to graph this solution on \([0.75, 5]\). [Hint. While the range for \(x\) is \([0.75, 5]\), the range for \(t\) should be between \(-3\) and \(25\) or so in order to get the full picture. Take this into account when you build your meshgrid.]
As a check, solve the equation from part (a) and get an explicit formula for \(x\), then graph it on \([0.75, 5]\). [hopefully they match!]
Check that solutions to the equation \[\frac{\dx}{\dt} = \frac{e^{2x}}{1 + t^2}\] satisfy \(e^{-2x} = -2\arctan(t) + c\).
What values of the constant \(c\) correspond to the initial conditions \(x(0) = 0\), \(\ln(2)\), and \(1\)?
Produce a contour plot of the solutions of the three initial value problems considered in part (b). Plot them on \([0, 1]\). [Hint. The vertical range should be \([0, 5]\) or so. The inverse tangent function is implemented in MATLAB by the command atan.]
For problems #17–19, perform the following tasks. (a) Separate the differential equation to get an implicit relationship that \(x\) and \(t\) satisfy, then (b) determine what values of the constant of integration will give the requested initial conditions, and finally (c) plot those solutions on the ranges of \(t\) and \(x\) specified.
\(\displaystyle \frac{\dx}{\dt} = \frac{t\cos(t)}{1 + 2x}\), \(x(0) = 1\), \(\frac{3}{2}\), \(2\), \(x \in [-2, 2]\), \(t \in [-5, 5]\)
\(\displaystyle \frac{\dx}{\dt} = \frac{x (\log(x))^2}{t^2}\), \(x(1) = e\), \(e^2\), \(e^3\) \(x \in [-3, 1]\), \(t \in [0, 5]\)
\(\displaystyle \frac{\dx}{\dt} = \frac{4tx}{1+x} \), \(x(1) = 1\), \(3\), \(5\) \(x \in [-3, 3]\), \(t \in [0, 10]\)
Plot a direction field for the equation \(y' = cos(x+y)\) on \([-2, 2] \times [-2, 2]\).
Plot a direction field for the differential equation \(y' = -\frac{t}{y}\) on \([-5, 5] \times [-5, 5]\). The directions in the field should seem to rotate around the origin if you have the correct picture; how do you know that’s the right behavior? [Note. If you want practice drawing direction fields by hand, this is a good one to work on.]
Consider the differential equation \[\frac{\dy}{\dx} = \frac{2 \cos(x) - 3 y}{x + y + 1} \,.\]
Where could solutions of this equation fail to exist, or fail to be unique?
Plot the direction field for this equation on \([0, 10] \times [-5, 5]\). Do you see some evidence supporting your answer to part (a)?
Let \(f(x)\) be the solution to the initial value problem \(y' = e^{-x^2}\), \(y(0) = 0\). [This function is essentially the error function, up to a constant.] Is \(\lim_{x \to \infty} f(x)\) finite or is it infinite? Explain how the direction field of the differential equation offers evidence of this.
\(\displaystyle \frac{\dx}{\dt} = x^2-\mu\,x\)
Is the stability different if \(\mu\) changes sign? Plot the phase lines for \(\mu<0\) and \(\mu >0\).
\(\displaystyle \frac{\du}{\dt} = (5-u)^{\mu}\)
Find \(u(t)\) as \(t\rightarrow\infty\). For what values of \(\mu\) is it different?