| Math 351: ODEs | Fall 2017 | |
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Unit One
Homework Help |
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Section 1.1
| A2. | The independent variable is $x$. The dependent variable is $y$. The equation is linear. |
| A6. | The independent variable is $t$. The dependent variable is $r$. The equation is linear, 2nd order. |
| A13b. | The independent variable is $t$. The system is linear. |
| B1. | The terms $\displaystyle{(a^2 - a)x\frac{dx}{dt}}$ and $te^{(a-1)x}$ are non-linear. The other terms are linear. The equation is linear if $a^2 - a = 0$ and $a-1=0$. |
Section 1.2
| A8. |
Here is a "short-hand" way to approach this exercise. This will make more sense, perhaps, if you consult a calc text such as Thomas' Calculus 12/e or 13/e. The hyperbolic sine ($\sinh x$) and hyperbolic cosine ($\cosh x$) are defined in the following way:
$\displaystyle{\sinh x = \frac{e^x - e^{-x}}{2}}$ and $\displaystyle{\cosh x = \frac{e^x + e^{-x}}{2}}$
Note that $\frac{d}{dx}\sinh x = \cosh x$ and $\frac{d}{dx}\cosh x = \sinh x$ A very helpful identity is $\cosh^2 x - \sinh^2 x = 1$. Now, the $y$ defined in this exercise is actually $y = \frac{1}{a}\cosh (ax)$, so that $y'' = a \cosh (ax)$, and $(y')^2 = \sinh^2 (ax)$ |
| A9. | The "T" in the denominator is supposed to be lower case. That is, $\displaystyle{y=\int_1^x \frac{\sin t}{t} dt}$. |
| B6. |
(a) For $x$ in the interval $(0,\infty)$, $\ln(|C_1x|)=\ln(|C_1|x)$. The term $|C_1|$ will cancel in the derivative expression $y'$.
(b) By the inverse function relationship of $\ln$ and $e$, the constant $C_2$ can be expressed as $C_2=\ln e^{C_2}$. Next, use a property of logarithms to find an expression for the single parameter $C$. |
| B8. | Assume $y(x) = ax^2 + bx + c$. then $y'(x) = 2ax + b$. Determine values for $a, b$ and $c$ through the differential equation by subbing for $y$ and $y`$ and equating coefficients of like powers of $x$. You should find that $a = -3$, etc. |
| P1-1. | |
| a. | Note that $\frac{dy}{dt}$ is zero for $y=0$ and $y=1$. The derivative is negative for $y$ less than zero, positive for $0 \lt y \lt 1$, and negative for $y \gt 1$ |
| b. | Use the product rule on the formula for $\displaystyle{\frac{dy}{dt}}$ to find that $\displaystyle{\frac{d^2 y}{d t^2} =y'(1-2y)}$. Sub the formula for $y'$ to get the full, explixcit formula for $y''$. |
| c. | The graph should increase, be concave down, and have $y=1$ as a horizontal asymptote. |
| d. | The graph should decrease, be concave up, and have $y=1$ as a horizontal asymptote. |
| e. | $y=1$ is a constant solution. |
| f. | Think of the limiting values in each case. |
Section 1.3
| A1. | There is a typo in the answer in the back of the book. The answer should be $-t(1+\cos t)$. |
| B1. | Find the integral curve to the first order IVP: $\displaystyle{\frac{dv}{dt} = \frac{1}{t^2 + 1} , t \ge 0, v(0) = 0}$. Then determine the limit of $v(t)$ as $t \rightarrow \infty$. |
| B8. |
a) The IVP for this case is: $x''(t) = a$ with $x'(0) = 0$ and $x(0) = 0$. Use the fact that $x'(3.8/3600) = 62$ to determine $a$. Then, determine the time $t^*$ at which the car is traveling at 60 mph from the $at^* = 60$. Be sure not to mix units. Perhaps it is best to use hours for time, so that the time for the car to go from 0 to 62 mph is 3.8/3600 hours. The final answer for this part is 0.03064 miles or 161.81 feet.
b) The simplest IVP for this part is $x''(t) = -d$ with $x'(0) = 62$ and $x(0) = 0$. Be sure to change feet to miles and seconds to hours if you work in mph. If t* is the time required to stop, then the acceleration is -62/t*. The position function is $\displaystyle{x(t) = \frac{-62}{t^*}\frac{t^2}{2} + 62t}$. The time required to stop is 0.00696 hours or 2.507 seconds. |
| B11. |
(a) $u(t) = Ae^{kat}$.
(b) Begin by subbing the expression for $u$ found in part (a) into the differential equation for $w$. Then, consider two cases. Case 1: $k=0$ You should find that $w(t) = aAt$. Case 2: $0 \lt k \le 1$ Here, $\displaystyle{w(t) = \frac{(k-1)A}{k}(1-e^{kat})}$ |
Section 2.1
| A2. | Note that $\displaystyle{\frac{x}{x+1} = \frac{(x+1) - 1}{x+1} = 1 - \frac{1}{x+1}}$. The answer is $y(x) = C(x+1)e^{-x}$. |
| A4. |
$g(y) = (y-1)(y-2)$. One of the zeros of $g$ is a singular solution. Use partial fractions to integrate $1/g(y)$. That is, set $\displaystyle{\frac{1}{(y-1)(y-2)} = \frac{A}{y-1} + \frac{B}{y-2}}$. At one point you should find something like $\displaystyle{\left|\frac{y-2}{y-1}\right|=C|x|}$.
You can transform this implicit expression for $y$ into an explicit function using an assumption on $x$ and some analysis on the positive or negative sign of the rational expression involving $y$. First, $x$ must be greater than or less than zero for the ODE to be defined, so assume $x$ is positive. Now you have $\displaystyle{\left|\frac{y-2}{y-1}\right|=Cx}$. Next, $y = 1$ and $y = 2$ are constant solutions. Other values of $y$ determine solution curves between these (horizontal) constant solutions. Therefore, the value of $\displaystyle{\frac{y-2}{y-1}}$ will be either positive or negative, but not both, for solution curves other than the constant solutions. That sign (positive or negative) can be "accounted for" in the constant $C$. Consequently, it is correct to write that $\displaystyle{\frac{y-2}{y-1}=Cx}$, and solve this expression for explicit formula for $y$. |
| A7. | After a proper separation of variables, use long division to find $\displaystyle{\frac{y^2}{y-1} = y + 1 + \frac{1}{y-1}}$. |
| A10. | $y' = 1 + x + (1 + x)y^2 = (1 + x)(1 + y^2)$. The solution for $y$ involves the tangent function. Refer to section 1 of the Key Points handout. |
| A12. | Let $z = y + 2x - 3$ to get the separated equation $\displaystyle{\frac{dz}{z+2}=dx}$. Integrate, exponentiate to find an explicit expression for $z$, and back substitute to give an explicit formula for $x$. This formula should have three terms, $e^x$, a term for $x$, and a constant term. |
| A13. | Following the suggested substitution, let $z=x+2y$ so that the original ODE is transformed to $\displaystyle{\frac{dz}{dx} = \frac{z+2}{z}}$. Note that $z=-2$ is a constant solution for this ODE, so it represents a possible singular solution. To solve the ODE in $z$, after separation, write $\displaystyle{\frac{z}{z+2} = 1 - \frac{2}{z+2}}$ (done by long division) before anti-differentiating. The singular solution $z=-2$, which translates to $x+2y = -2$ is the one that satisfies the given IC of $y(0)=-1$. |
| A15. | The implicit solution is sufficient. That is, there is no need to find the explicit solution. |
| C1. |
The differential equation for velocity $v$ is $\frac{dv}{dt} = k\sqrt{v}$. Conditions on the velocity are given at $t=0$ and $t = 0.1$, so this is a boundary value problem. $v(0) = V_0$ and $v(0.1) = 0$. Note that we are trying to find the value of $V_0$. The two conditions on $V$ enable us to determine the constant of integration $C$ and the parameter $k$ in terms of the initial velocity $V_0$.
Your solution for $v(t)$ should be $v(t) = [-10\sqrt{V_0}t + \sqrt{V_0}\;]^2$. There is a second BVP on the position function: $\frac{dx}{dt} = v(t)$, where $v(t)$ is the formula determined in the first step outlined above. The boundary conditions on $x$: $x(0) = 0$ and $x(0.1) = 10$. From this BVP solution one can determine the value for $V_0$. |
| C2. | The ODE is separable, of course, and is algebraically reducible to a simple form by dividing the denominator into the two terms in the numerator. For initial conditions, assume $L(0) = 0$. Do not solve the result for $L$. Rather, express $t$ as a function of $L$. |
| C4. |
a) Take $C(0)$ to be 14 mg/L. The solution is $C(t) = 14e^{-t/6}$.
b) Determine the time $t$ at which the blood concentration is 5 mg/L. c) The second injection is given at $t$=6, so the concentration remaining from the first injection is $14e^{-6/6} = 14/e$. The second injection has an initial concentration of 14 at this time. Now take $t=0$ to be the time of the second injection, so that the "initial" concentration is $14/e + 14$. The blood concentration for $t$ bigger than zero is $C(t) = (14/e + 14)e^{-t/6}$. d) Solve $14e^{-t/6} = 6$. |
| C6. |
In case you had not noticed, this ODE is that of Project 1-1. IN this exercise, analytic techniques will be used to give answers to similar questions posed in the project, where qualitative methods were used.
a) You should find that $\displaystyle{P(t) = \frac{Ce^t}{1 + Ce^t}}$ You may leave your answer in this form. b) Using $P(0) = P_0$ you should find that $\displaystyle{P(t) = \frac{P_0e^t}{(1 - P_0) + P_0e^t}}$. Divide all terms by $e^t$ will facilitate finding the limit for this case and that in (c). |
Section 2.2
| A2. | The integration factor here is $e^{x^2}$. The answer is $y(x)=\left(\frac{x^2}{2} + C\right)e^{-x^2}$. |
| A4. | The integration factor here is $e^x$. Use integration by parts twice on $\displaystyle{\int e^x \cos x dx }$. The general solution is $y(x) = \frac{1}{2}(\sin x + \cos x) + Ce^{-x}$. |
| A14. | The integration factor is $\displaystyle{\mu(x) = \frac{e^{-1/x}}{x^2}}$. Integrate $\displaystyle{\int\frac{e^{-1/x}}{x^2} dx }$ using $u$ substitution with $u = -1/x$. The general solution is $y(x) = x^2 + Cx^2e^{1/x}$. |
| B1. | Dividing both sides by $y^2$ gives $y^{-2}y'+\left(\frac{-4}{t}\right)y^{-1}=-6t$. Letting $z=y^{-1}$ means $\frac{dz}{dt}=-y^{-2}\cdot\frac{dy}{dt}$. The new linear ODE in $z$ has an integrating factor of $t^4$. |
| B8. | Show that the linear DE $\displaystyle{a_1(x)\frac{dy}{dx} + a_0(x)y = 0}$ can be written in the form $\frac{dy}{dx} = h(x)g(y)$. |
| B13. | (a) $\displaystyle{p(t) = \frac{\nu}{\mu + \nu}\bigg[ 1 - e^{-(\mu + \nu)t} \; \; \bigg] + p_0e^{-(\mu + \nu)t}}$ and $\displaystyle{q(t) = \frac{\mu}{\mu + \nu}\bigg[ 1 - e^{-(\mu + \nu)t}\;\;\bigg] + q_0e^{-(\mu + \nu)t}}$ |
Section 2.3
| B2. | The volume in the tank as a function of time is 50 + t, so the outflow concentration is $\frac{q(t)}{50 + t}$. Answer: 9.375 grams/gallon |
| B4. |
a) A 1% concentration means 1 gallon of chlorine per 100 gallons of solution. The answer is 0.75% solution.
b) This part may be solved without using a differential equation. Answer: 0.00875 gm/gal = 0.875% solution. |
| B8. | The differential equation is $\frac{dW}{dt} = 50 - \frac{100W}{6000-50t}$, where $W(t)$ is the number of women on staff at time $t$ measured in weeks. Answer: 50%. |
| C1. | The governing ODE is $\frac{dX}{dt} + \frac{rX}{V} = rc$, where $X$ is the quantity of drugs in the organ at time $t$. The quantity, when the concentration is $c_{max}$ , is given by $V \cdot c_{max}$ |
Section 2.4
| A2, A3, and A8. | Sketch these slope fields by hand using isoclines. I encourage you to check your slope field sketches using Sage. There is a published Sage worksheet entitled "ODE_dirField_v3" that has a group of cells for generating a slope field. |
| B1. | Generate the slope field using either Sage and sketch it on your paper if you have trouble sketching this slope field using isoclines. There is a published Sage worksheet entitled "ODE_dirField_v3" that has a group of cells for generating a slope field. |
| C1. | We are given that $\phi'(t) = f(\phi(t))$. We want to show that $\frac{d}{dt}\phi(t + k) = f(\phi(t+k))$. Take the derivative of $\phi(t+k)$ using the chain rule. That is, $\frac{d}{dt}\phi(t + k) = \phi'(t+k)\frac{d}{dt}(t+k)$. Now, make appropriate substitutions for the two terms on the RHS. |
Section 2.5
| A8. | There are no equilibrium solutions for $f(x) = 1 - \frac{x}{1+x}$. Note that $\frac{x}{1+x}$ has a limiting value of "1" as x goes to $\pm \infty$. |
| B5. | b. Remember that $P(0)= 3$ as given in part a. |
| B8. | A rational function in $f(Q) = \frac{p(Q)}{r(Q)}$, where $p(Q)$ and $r(Q)$ are polynomials, is the easiest formula to construct that matches the indicated behavior. $p(Q)$ will be zero at the equilibrium solutions of the ODE. Note that the direction arrow change between -1 and 2 with NO equilibrium solution shown. How else can the direction change without the derivative going to zero? |
Section 2.6
| A4. | There are three equilibrium points, one of each type. |
| B1. |
(a) Solve $\frac{8P}{u} - bSu^2 = 0$ for $u$.
(b) When $u$ is a very small, positive number (less than the equilibrium solution), the term $\frac{8P}{u}$ is very large and positive, and the term $bSu^2$ is very small, so $\frac{du}{dt}$ is positive. What about the case when $u$ is a very large positive number (greater than the equilibrium solution)? |
| B3. |
I would do part (c) first.
(c) There is one ES, $N = 1/b$. (d) The inflection point is $N = \frac{1}{b \cdot e}$, where $e$ is the natural exponential base. This value can be determined by finding $\frac{d^2N}{dt^2}$ using the chain rule. That is, you should find that $\frac{d^2N}{dt^2} = \left[-a\ln(bN) - a\right]\frac{dN}{dt}$. The inflection point occurs where the second derivative is zero. |
Section 2.7
| A2. |
a) The graph of $f(x)$ is a parabola that opens up, so there are as many as three qualitatively different graphs for $f$: one with no ES, one with exactly one, and one with two.
b) There are two bifurcation values for $c$. c) When there are two ES for a given $c$, one is a source and the other is a sink. |
| A3. | a) $x$=0 is an equilibrium solution for any $c$. Note that the graph of $f$ is qualitatively different for $c < 0$, $c = 0$, $0 < c < 1$, $c =1$, and $c > 1$. I disagree with the author on the number of bifurcations. I say there are two (not just $c$=1). |
| B1. | The graph of $f(P)$ is a parabola that opens downward. For what $h$ is there no longer any equilibrium solutions? |
| C1. | (c) The graph you sketch is a bifucation diagram. It will look very similar to Figure 2.27 in the text. |
Section 2.8
| A4. | Find the largest rectangle centered on the point $(x_0, y_0)$. |
| A9. | You will have to find the solution for the IVP to determine the largest interval. The equation is separable. The answer involves the tangent function which is not defined for all real $t$. |
| B3. | Of course one must consider the partial of $f(x,y)$ with respect to $y$ for the case of uniqueness. |
| C6. | This question has less to do with the existence and uniqueness statement and more to do with the modeling differential equation accurately representing reality, the chemical reaction in this case. If something can't happen in a mathematical sense (the modeling ODE does not have a solution) does it imply it cannot happen in a "real" sense (the reaction cannot happen)? How about if the mathematics suggests something will happen (the model ODE has a solution), does that imply the modeled reaction must happen in the "real" world? |
Section 3.1
| A2. | Four steps are required for the stepsize of 0.5. |
| A8. |
Just report the value for $y(4)$ determined using Sage. You do not need to turn in the Sage worksheet.
Sage will default to symbolic and rational mode if not "forced" into floating point mode. The later mode is much more efficient for numerical schemes. You can force Sage to work in floating point mode by using floats (decimals) for the intial values and step size. For example, use x_0 = 0.0 instead of 0, y_0 = 1.0 instead of 1. Also, when you define your functions, use floating point numbers and approximations for number such as $\pi$. In this problem, the function should be defined as f(x, y) = y * sin(3.0 * x). |
| B3. | Estimate $V(0)$ using Sage and report it on your paper. See the comments for exercise A8 above. |
| C2. |
(b) Your Euler results for $h$ values of 0.1 and 0.05 should NOT be very good. That is, your value for $y(1)$ will be HUGE (like $10^{10}$) for both these $h$ values.
(c) Generate a plot for the case $h=0.01$ only. The other two cases would have only a single point for the Euler solution.. |
Section 3.2
| A3. |
Section 3.3
| A2. |
Section 3.4