7.6 Exercises
Exercise 7.1 Determine the value of \(C\) when \(I(0)=10\) for the two equations:
\[\begin{equation} \begin{split} I_{1}(t) = 1000 + Ce^{-.03t} \\ I_{2}(t) = 1000 + C e^{-0.015 t^{2}} \end{split} \end{equation}\]
Exercise 7.3 A chemical reaction \(2A \rightarrow C + D\) can be modeled with the following differential equation (Scholz and Scholz 2014):
\[\begin{equation} \frac{dA}{dt} = -2 k A^{2} \end{equation}\]
Apply the method of separation of variables to determine a general solution for this differential equation.
Exercise 7.4 Which of the following differential equations be solved via separation of variables?
- \(\displaystyle \frac{dy}{dx} = x \cdot (y^{2}+2)\)
- \(\displaystyle \frac{dy}{dx} = x^{2} + xy\)
- \(\displaystyle \frac{dy}{dx} = e^{x+y}\)
- \(\displaystyle \frac{dy}{dx} = y \cdot \cos(2+x)\)
- \(\displaystyle \frac{dy}{dx} = \ln x + \ln y\)
Exercise 7.5 Solve the following differential equations by separation of variables:
- \(\displaystyle \frac{dy}{dx} = \frac{y^{3}}{x}\)
- \(\displaystyle \frac{dy}{dx} = 1+y^{2}\)
- \(\displaystyle \frac{dy}{dx} = 8-y\)
Exercise 7.6 Consider the following differential equation \(\displaystyle \frac{dP}{dt} = - \delta P\), \(P(0)=P_{0}\), where \(\delta\) is a constant parameter.
- Solve this equation using the method of separation of variables.
- Solve this euqation using an integrating factor.
- Your two solutions from the two methods should be the same - are they?
Exercise 7.7 Here we return to the problem of how animals consume food. A differential equation that relates a consumer’s nutrient content (denoted as \(y\)) to the nutrient content of food (denoted as \(x\)) is given by:
\[\begin{equation} \frac{dy}{dx} = \frac{1}{\theta} \frac{y}{x}, \end{equation}\]
where \(\theta \geq 1\) is a constant. Apply separation of variables to determine the general solution to this differential equation.
Exercise 7.9 A plant grows propritional to its current length \(L\). Assume this proportionality constant is \(\mu\), whose rate also decreases proportional to its current value. The system of equations that models this plant growth is the following: \[\begin{equation} \begin{split} \frac{dL}{dt} = \mu L \\ \frac{d\mu}{dt} = -k \mu \\ \mbox{($k$ is a constant parameter)} \end{split} \end{equation}\] Apply separation of variables to determine the general solutions to this system of equations.
Exercise 7.10 Use the method developed in this section determine the general solution to the following system of differential equations: \[\begin{equation} \begin{split} \frac{dx}{dt} &= x-y \\ \frac{dy}{dt} & = 2y \end{split} \end{equation}\]
Exercise 7.12 For each of the following differential equations:
- Determine equilibrium solutions for the differential equation.
- Apply separation of variables to determine general solutions to the following differential equations:
- Choose reasonable values of any parameters and plot the solution curve for an initial condition that you select.
- \(\displaystyle \frac{dy}{dx} = -\frac{x}{y}\)
- \(\displaystyle \frac{dy}{dx} = 8-y\)
- \(\displaystyle \frac{dW}{dt} = k (N-W)\) (\(k\) and \(N\) are constant parameters)
- \(\displaystyle \frac{dR}{dt} =-aR \ln \frac{R}{K}\) (\(a\) and \(K\) are constant parameters)
Exercise 7.14 An alternative model of mayfly mortality is the following: \[\begin{equation} \displaystyle \frac{dM}{dt} = - \delta(t) \cdot M, \end{equation}\] where \(\delta(t)\) is a time dependent mortality function. Determine a solution and plot a solution curve (assuming \(M(0)=10,000\) and over the interval from \(0 \leq t \leq 1\)) for this differential equation when \(\delta(t)\) has the following forms:
- \(\delta(t) = t^{2}\)
- \(\delta(t) = 1-t^{2}\)