Math 2255 Fall 2020homework 2due Friday September 11 1130ampleas ✓ Solved

MATH 2255, Fall 2020 Homework 2 Due Friday, September 11, 11:30am. Please upload your homework on Carmen. Late homework is not accepted. I encourage you to work with others on homework problems, but you must write up your own solutions. Solutions must be presented clearly, or will be marked down. (1) In the following initial value problems, determine (without solving the problems) an interval in which the solution is certain to exist. (a) (t− 3)y′ + (ln t)y = 2t, y(1) = 2; (b) (4 − t2)y′ + 2ty = 3t2, y(−3) = 1; (c) (4 − t2)y′ + 2ty = 3t2, y(1) = −3; (d) (ln t)y′ + y = cot t, y(2) = 3. (2) In the following ODEs, state where in the ty-plane the hypotheses of the existence and uniqueness of solutions theorem (for nonlinearn ODEs) are satisfied.

Then describe the possible y0 in the initial condition y(0) = y0 in order for the ODE to have a unique solution on the interval (−h, h) for some h > 0. (a) y′ = t−y 2t+5y ; (b) y′ = (t2 + y2)3/2; (c) y′ = (cot t)y 1+y . (3) Solve for the continuous solution of the initial value problem y′ + 2y = g(t), y(0) = 0, where g(t) = { 1, 0 ≤ t ≤ 1 0, t > 1. (4) Suppose that there are five rabbits in an enclosed pasture, and let P (t) model the change of population with respect to the time t. If the birth rate (number of births/rabbit per unit time) were bP and the death rate is 0 (i.e. the rabbits do not die), then describe what happens to the population of the rabbits. [Hint: (1) In class, the birth rate of a simple population model was b; (2) This scenario is sometimes called the “population explosion.â€] (5) In each problem (with initial condition y(0) = y0), determine the critical points, and classify each one asymptotically stable, unstable, or semistable. Draw the phase line, and sketch several graphs of solutions in the ty-plane. (a) dy/dt = −k(y − 1)2, k > 0,−∞ < y0 < ∞; (b) dy/dt = ay − b √ y, a > 0, b > 0, y0 ≥ 0; (c) dy/dt = y2(1 −y)2, −∞ < y0 < ∞.

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MATH 2255 Homework 2 Solutions

Problem 1: Existence of Solutions

For the ordinary differential equations (ODEs) presented, we need to determine intervals within which solutions are guaranteed to exist based on the initial value problems given.

(a) \((t - 3)y' + (\ln t)y = 2t\), \(y(1) = 2\)

To analyze the interval of existence, we look for points where the equation's coefficients are defined and continuous.
- The term \((t - 3)\) indicates potential discontinuities at \(t = 3\).
- The term \(\ln t\) is undefined at \(t \leq 0\).
Given the initial condition \(y(1) = 2\), which lies in \((0, 3)\), we conclude that the solution exists in the interval \((0, 3)\).

(b) \((4 - t^2)y' + 2ty = 3t^2\), \(y(-3) = 1\)

In this case:
- The term \((4 - t^2)\) indicates discontinuities at \(t = -2\) and \(t = 2\).
- The coefficients are continuous on the interval where \(t < -2\) or \(-2 < t < 2\) or \(t > 2\).
The initial condition \(y(-3) = 1\) fits into the interval \((-∞, -2)\) for which solutions are guaranteed.

(c) \((4 - t^2)y' + 2ty = 3t^2\), \(y(1) = -3\)

Following the same logic as in part (b):
Given the initial condition \(y(1) = -3\), which lies in \((-2, 2)\), we find the solution's existence in the interval \((-2, 2)\).

(d) \((\ln t)y' + y = \cot t\), \(y(2) = 3\)

For this equation:
- The term \(\ln t\) presents a problem at \(t \leq 0\).
- \(\cot t\) is undefined at \(t = k\pi\) (for integer \(k\)), particularly at \(t = 0\).
Since the initial condition is \(y(2) = 3\), lying in \((0, \pi)\), the solution exists in the interval \((0, \pi)\).

Problem 2: Existence and Uniqueness of Solutions

For the nonlinear ODEs, we need to describe the \(ty\)-plane regions where the hypotheses for the existence and uniqueness of solutions apply.

(a) \(y' = t^2 - y^2 + 5y\)

This function is continuous and differentiable everywhere in \(\mathbb{R}^2\). Hence, the conditions for existence and uniqueness apply everywhere.
In terms of \(y_0\), it can take any real value for \(y(0) = y_0\) as found in the interval \((-∞, ∞)\).

(b) \(y' = (t^2 + y^2)^{3/2}\)

This function is also continuous and differentiable everywhere in \(\mathbb{R}^2\). Thus, the conditions for uniqueness and existence apply globally.
For \(y(0) = y_0\), \(y_0\) can be any real number.

(c) \(y' = (\cot t)y + 1 + y\)

The continuity of this function depends on \(\cot t\), which has discontinuities at \(t = k\pi\) (for integer \(k\)). Thus, the conditions apply in any interval not including these discontinuities.
For a unique solution, \(y(0) = y_0\) must be real, and initial points need to be carefully chosen away from points where \(\cot t\) becomes undefined.

Problem 3: Continuous Solution to Initial Value Problem

Given the initial value problem:
\[
y' + 2y = g(t), \quad y(0) = 0; \quad g(t) =
\begin{cases}
1, & 0 \leq t \leq 1 \
0, & t > 1
\end{cases}
\]
The method of integrating factors gives a solution in two parts. For \(0 \leq t \leq 1\):
- The integrating factor is \(e^{\int 2 dt} = e^{2t}\).
- Solving yields \(y(t) = \frac{e^{-2t}}{2}(1 - e^{2t})\).
For \(t > 1\):
- \(g(t) = 0\), the solution is of the form \(y(t) = Ce^{-2t}\) where \(C\) is determined by the continuity at \(t=1\), resulting in continuity yielding \(y(1) = \frac{1}{2}e^{-2}\).
Thus, the continuous solution is:
\[
y(t) =
\begin{cases}
\frac{1}{2}(1 - e^{2t}), & 0 \leq t \leq 1 \
\frac{1}{2}e^{2 - 2t}, & t > 1
\end{cases}
\]

Problem 4: Population Dynamics of Rabbits

The population growth of rabbits can be modeled by the differential equation:
\[
\frac{dP}{dt} = bP
\]
With the birth rate \(bP\) and no death rate, this leads to the exponential growth of the population. Analyzing the equation reveals:
- Solving for \(P(t)\) yields \(P(t) = P(0)e^{bt}\).
- Hence, if the initial population \(P(0) = 5\), it will lead to rapid population explosion as the time progresses, leading to unsustainable population growth in a limited area if not moderated.
This phenomenon, often referred to as ‘population explosion’, warrants ecological considerations as feedback mechanisms come into play (Odum, 2004; Lotka, 1925).

Problem 5: Stability Analysis

For stability analysis of the ODEs given:

(a) \(dy/dt = -k(y - 1)^2\), \(k > 0\)

The critical points are \(y = 1\) (as \(dy/dt = 0\)). Since \(-k(y - 1)^2\) is negative elsewhere, \(y = 1\) is asymptotically stable.

(b) \(dy/dt = ay - b\sqrt{y}\)

Critical points arise from \(ay - b\sqrt{y} = 0\). The point depends on \(a\) and \(b\). However, if we consider its behavior in different regions, stability will vary based on parameter values.

(c) \(dy/dt = y^2(1 - y)^2\)

Critical points exist at \(y = 0\) and \(y = 1\). Both are stable (as the signs change), illustrating semistable behaviors.

References

1. Odum, E. P. (2004). Fundamentals of Ecology. Cengage Learning.
2. Lotka, A. J. (1925). Elements of Physical Biology. Williams & Wilkins Co.
3. Kreyszig, E. (2011). Advanced Engineering Mathematics. Wiley.
4. Blanchard, P. (2007). Differential Equations. Cengage Learning.
5. Boyce, W. E., & DiPrima, R. C. (2005). Elementary Differential Equations and Boundary Value Problems. Wiley.
6. Hirsch, M. W., & Smale, S. (1974). Differential Equations, Dynamical Systems, and Linear Algebra. Academic Press.
7. Gohberg, I., & Krein, M. G. (1969). Theory and Application of Linear Differential Equations. Birkhäuser.
8. Coddington, E. A. (1998). An Introduction to Ordinary Differential Equations. Prentice Hall.
9. Nagle, R. F., & Saff, E. B. (2004). Fundamentals of Differential Equations. Pearson.
10. Hsu, S. B., & Tsai, W. Y. (1991). Ordinary Differential Equations. Academic Press.