\documentclass[12pt]{amsart} \setlength{\textheight}{9in} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{0in} \setlength{\evensidemargin}{0in} \setlength{\topmargin}{-0.5in} \usepackage{amssymb,amsmath,amsfonts,graphicx,psfrag} %\pagestyle{empty} \newcommand{\dis}{\displaystyle} \newcommand{\bd}{\partial} \begin{document} {\Large \begin{center} {\bf Exact First Order Differential Equations} \end{center} This Lecture covers material in Section 2.6. A first order differential equations is {\bf exact} if it can be written in the form \[ M(x,y) + N(x,y)\frac{dy}{dx} = 0, \] where \[ \frac{\bd M}{\bd y} = \frac{\bd N}{\bd x}. \] Before showing how to solve these we need to review some multivariable calculus, especially the {\bf two-variable chain rule}. This will also help to motivate why equations of this form are important in physics. Let $\psi(x,y)$ be a function of two variables. Then we can think of \[z = \psi(x,y) \] as a surface in three-dimensional space where $z$ is the height above the $xy$-plane. Now suppose the $x$ and $y$ are functions of $t$ (time) so that $( x(t), y(t) )$ gives a curve in the $xy$-plane. Then $z(t) = \psi(x(t),y(t))$ gives a curve in three-dimensional space. Suppose we desire to know the rate of change of $z$ with respect to $t$. According to the two-variable chain rule the answer is \[ \frac{dz}{dt} = \frac{\bd \psi}{\bd x}\frac{dx}{dt} + \frac{\bd \psi}{\bd y}\frac{dy}{dt}. \hspace{1in} (*) \] This formula is derived in Calculus III. Here I will give an intuitive motivation for why it works. Suppose $\psi(x,y)$ is just a plane. Then we have \[ z = \psi(x,y) = Ax + By + C \] for some constants $A$, $B$ and $C$. Here $A$ is the slope of the plane with respect to the $x$ direction, $B$ is the slope of the plane with respect to the $y$ direction and $C$ is the intercept with the $z$-axis. \begin{center} \includegraphics[height=2in]{Figs/planeinR3.pdf} $z = -\frac{1}{2}x-\frac{3}{4}y+3$ \end{center} \vspace{.25in} We want to compute the change in $z$ as $t$ changes from $t_0$ to $t_0 + \Delta t$. Let $\Delta x = x(t_0 + \Delta t) - x(t_0)$ and $\Delta y = y(t_0 + \Delta t) - y(t_0)$ be the changes in $x$ and $y$, respectively. For convenience let $x_0 = x(t_0)$ and $y_0 = y(t_0)$. Then the change in $z$ is \[ \Delta z = \psi(x_0+\Delta x, y_0 + \Delta y) - \psi(x_0, y_0) = A\Delta x + B \Delta y. \] We divide both sides by $\Delta t$ to obtain \[ \frac{\Delta z}{\Delta t} = A \frac{\Delta x}{\Delta t} + B \frac{\Delta y}{\Delta t}. \] Now we can find the derivative of $z$ with respect to $t$ by taking limits as $\Delta t \to 0$. This gives \[ \frac{dz}{dt} = A\frac{dx}{dt} + B \frac{dy}{dt}. \] But notice that $A = \bd_x \psi$ and $B=\bd_y \psi$. Thus we have \[ \frac{dz}{dt} = \frac{\bd \psi}{\bd x}\frac{dx}{dt} + \frac{\bd \psi}{\bd y}\frac{dy}{dt} \] which is $(*)$. This shows that the two-variable chain rule works for planes. In general if $\psi(x,y)$ is reasonably smooth it can be approximated near each point by a tangent plane. It can be shown that this gives the two-variable chain rule for any function of two variables that is smooth enough that its graph has a tangent plane at each point in an open set containing the point of interest. Now, let $z=\psi(x,y)$ be a surface. But suppose $z$ is some quantity that is conserved, like energy. That is we now have \[ \psi(x,y) = C. \] The slice of the surface through $z=C$ is called {\bf level curve}. Example: Let $z = \psi(x,y) = x^2 + y^2$. Then the level curve for $z=1$ is a circle of radius 1 that floats one unit above the $xy$-plane. Example: Let $z = \psi(x,y) = 3x+y-3$. The level curve for $z=2$ is the line $3x+y-3 = 2$, or $y=-3x + 5$, that is it floating two units above the $xy$-plane. For now suppose $y$ is a function of $x$ (at least implicitly). As we change $x$ we cause $y$ to change so that $z=\psi(x,y)$ stays on the same level curve. Since $z$ is not changing we have $dz/dx = 0$. The two-variable chain rule, using $x$ for $t$, gives \[ 0 = \frac{dz}{dx} = \frac{\psi(x,y(x))}{dx} = \frac{\bd \psi}{\bd x}\frac{dx}{dx} + \frac{\bd \psi}{\bd y}\frac{dy}{dx}. \] Therefore, \[ \frac{\bd \psi}{\bd x} + \frac{\bd \psi}{\bd y} y' = 0. \] If $\psi_x$ and $\psi_y$ are known functions what we have is a differential equation in $y$. We will be doing the inverse of this process. That is, given at differential equation in the form \[ M(x,y) + N(x,y) \frac{dy}{dx} = 0 \] we will solve it for $y(x)$ (or at least a relation between $x$ and $y$) by finding a surface $\psi(x,y)$ such that $M= \psi_x$ and $N=\psi_y$, and then using an initial condition to find the desired level curve. In many applications $\left< M, N \right>$ is given as a force field and then $\psi$ is a potential energy function. If energy is conserved, the dynamics are restricted to a level curve of $z=\psi(x,y)$. Enough talk, let's do some examples. {\bf Example 1.} Solve $(2x+y) + (x+2y)y'=0$, with $y(3)=1$. \begin{proof}[Solution] We want to find a function $\psi(x,y)$ such that \[ \frac{\bd \psi}{\bd x} = 2x+y \hspace{.5in}\&\hspace{0.5in}\frac{\bd \psi}{\bd y} = x+2y. \] So, we integrate. \[ \psi = \int \psi_x \, dx = \int 2x+y \, dx = x^2 + xy + C_1(y),\] where $C_1(y)$ an arbitrary function of $y$. The idea is we are finding the class of all functions whose partial derivative with respect to $x$ gives $2x+y$. But we also need for $\psi_y = x+2y$. So, we integrate. \[ \psi = \int \psi_y \, dy = \int x+2y \, dy = xy + y^2 + C_2(x),\] where $C_2(x)$ can be any function of $x$. We now have two classes of functions, each satisfying one of the two conditions. If we could find a function that is in both class that would do the trick. The answer is obvious. Let \[ \psi(x,y) = x^2 + xy + y^2. \] This function is in both classes and thus satisfies both the needed conditions. Now we consider the initial condition, $y(3)=1$, that is, $x=3$ $\implies$ $y=1$. Then \[ \psi(3,1) = 9+3+1=13. \] Thus, the level curve we want is \[ x^2 +xy + y^2 = 13. \] We will leave as a relation. Below are plots of the surface $z=x^2 +xy + y^2$ with the level 13 curve and a projection of this curve into the $xy$-plane. \end{proof} \begin{center} \includegraphics[width=2in]{Figs/surface1.jpg} \hspace{0.5in} \includegraphics[width=2in]{Figs/levelcurve1.jpg} \end{center} {\bf Extra Credit.} Prove that this curve $x^2+xy+y^2=13$ is an ellipse and find its focal points. You can do this by reviewing how to rotate graphs with rotation matrices and the properties of ellipses. Then rotate the graph $45^o$ so that its major axis lies along the x-axis. \vspace{.25in} {\bf Example 2 (Not!).} Solve $(2x+2y) + (x+2y)y'=0$, with $y(3)=1$. We integrate. \[ \psi = \int 2x+2y \, dx = x^2 + 2xy + C_1(y). \] \[ \psi = \int x+2y \, dy = xy + y^2 + C_2(x). \] Now look closely. Since $2xy \neq xy$ there is no function that meets both conditions. The method fails! What this means in physical terms is that the force field $\left< 2x+2y, x+2y \right>$ does not arise from a potential function; in such a system energy is {\bf not conserved}. \vspace{.1in} Note: This example can be converted to a separable equation because it is homogeneous. We work through this in the Appendix at the end of these notes. It is optional reading. \vspace{.1in} What we need is a quick test to see if $\psi$ exists for a given equation so that we don't waste a lot of time barking up the wrong tree. \vspace{.1in} {\bf Theorem!} Given two functions $M(x,y)$ and $N(x,y)$, there exists a function $\psi(x,y)$ such that \[ \frac{\bd \psi}{\bd x} = M(x,y) \hspace{0.5in}\&\hspace{0.5in} \frac{\bd \psi}{\bd y} = N(x,y), \] if and only if \[ \frac{\bd M}{\bd y} = \frac{\bd N}{\bd x}, \] in an open rectangle containing the point of interest. Check this for the two examples above. This is Theorem 2.6.1 in your textbook. Your textbook gives a proof, but another prove is covered in Calculus III that uses Green's Theorem. If you are a Math major read both and compare them. However, one direction is easy: if $\psi$ exists, then $\psi_{xy} = \psi_{yx} \implies M_y = N_x$. This theorem is the motivation for the definition we gave at the beginning of an exact first order differential equation. \vspace{.25in} {\bf Example 3.} Find the general solution to \[ y \cos x + y e^{xy} + (\sin x + x e^{xy}) y' = 0. \] \begin{proof}[Solution] Let $M= y \cos x + y e^{xy}$ and $N=\sin x + x e^{xy}$. Then \[ M_y = \cos x + e^{xy} + xy e^{xy} = N_x. \] Thus, it is exact. We integrate. \[ \psi = \int M \, dx = y \sin x + e^{xy} + C_1(y) \] and \[ \psi = \int N \, dy = y \sin x + e^{xy} + C_2(x). \] We let $\psi(x,y) = y \sin x + e^{xy}$. The general solution is then \[ y\sin x + e^{xy} = C. \] \end{proof} {\bf Example 4.} Solve $4x^3 + 4y^3 y' = 0$, with $y(1) = 1$. \begin{proof}[Solution] It is exact since $(4x^3)_y = 0$ and $(4y^3)_x = 0$. Then \[ \psi = \int 4x^3 \, dx = x^4 + C_1(y) \] and \[ \psi = \int 4y^3 \, dy = y^4 + C_2(x). \] We let $\psi = x^4 + y^4$. Since (1,1) is our initial condition we see that our solution is \[ x^4 + y^4 = 2. \] Below is a graph of this curve projected into the $xy$-plane. \end{proof} \begin{center} \includegraphics[width=2in]{Figs/x4y4.jpg} \end{center} {\bf Example 5.} Solve $4x^4 + 4xy^3 y' = 0$, with $y(1) = 1$. \begin{proof}[Solution] We check for exactness. $(4x^4)_y = 0$ while $(4xy^3)_x = 4y^3$. Thus it is not exact. But wait! Notice that this example is exactly the same as Example 4, but that we have multiplied through by $x$. So, if we now multiple through by $\frac{1}{x}$ we get \[ 4x^3 + 4y^3 y' = 0. \] Thus, the solution is same as in Example 4! \end{proof} \vspace{.25in} {\bf Integrating factors.} \vspace{.25in} This last example motivates the following idea. Suppose we have a differential equation of the form \[ M + N y' = 0 \] which is not exact. Can we find a function $\mu(x,y)$ such that \[ \mu M + \mu N y' = 0 \] is exact? The answer is, not always, but sometimes you can. When this works we call $\mu$ an {\bf integrating factor}. Finding such as $\mu$ can be tricky. Here we show three special cases where an integrating factor $\mu$ can be found. Each relies on an assumption about $\mu$ that can be tested for. \vspace{.25in} {\bf Case 1.} Suppose a suitable $\mu$ exists and that it is a function of $x$ only. \vspace{.25in} {\bf Case 2.} Suppose a suitable $\mu$ exists and that it is a function of $y$ only. \vspace{.25in} {\bf Case 3.} Suppose a suitable $\mu$ exists and that it can be written as a function dependent only on the product $xy$. \vspace{.25in} In all cases we need to find $\mu$ such that \[ (\mu M)_y = (\mu N)_x \] so that we have exactness. By the product rule this is equivalent to requiring \[ \mu_y M + \mu M_y = \mu_x N + \mu N_x. \hspace{1in} (*) \] \vspace{.25in} {\bf Case 1.} If Case 1 holds then $\mu_y =0$ and we can think of $\mu_x$ as $\mu'$. Then $(*)$ becomes \[ \mu M_y = \mu' N + \mu N_x \] or \[ \frac{\mu'}{\mu} = \frac{M_y - N_x}{N}. \] If our assumption is correct then, since $\mu'/\mu$ depends only on $x$, we know that $(M_y -N_x)/N$ depends only on $x$. Then the integrals below are well defined. \[ \int \frac{1}{\mu} \, d\mu = \int \frac{M_y - N_x}{N} \, dx \] Thus, \[ \mu = e^{\int \frac{M_y - N_x}{N} \, dx }. \] In fact, this gives us a test to determine when this method will work. If $ \frac{M_y - N_x}{N}$ depends only on $x$ it follows that $\mu$ depends only on $x$. \vspace{.25in} {\bf Example 6.} Find the general solution to $y^2 + x^3 + xy y' = 0$. \begin{proof}[Solution] Since $(y^2+x^3)_y = 2y$ and $(xy)_x = y$ are not equal, this equation is not exact. But \[ \frac{2y - y}{xy} = \frac{1}{x} \] depends only on $x$. Thus we let \[ \mu = e^{\int \frac{1}{x}\, dx} = x. \] So, we multiply through by $x$ to get \[ xy^2 + x^4 + x^2y y' = 0. \] Let $M=xy^2 + x^4$ and $N=x^2y$. Then $M_y =2xy = N_x$, so we have exactness. Now we find $\psi$ as before. \[ \psi = \int M \, dx = \frac{1}{2}x^2y^2 + \frac{1}{5}x^5 + C_1(y) \] \[ \psi = \int N\, dy = \frac{1}{2}x^2y^2 + C_2(x) \] Thus, we let $\psi = \frac{1}{2}x^2y^2 + \frac{1}{5}x^5$, so the general solution is \[ \frac{1}{2}x^2y^2 + \frac{1}{5}x^5 = C, \] or if you prefer \[ 5x ^2y^2 + 2x^5 = C. \] Solving for $y$ gives \[ y = \pm \sqrt{\frac{C-2x^5}{5x^2}}. \] \end{proof} \vspace{.25in} {\bf Case 2.} This is so similar to Case 1 that we leave it to you to develop the method and find the formula for $\mu(y)$. \vspace{.25in} {\bf Case 3.} Recall equation $(*)$: $\mu_y M + \mu M_y = \mu_x N + \mu N_x$. Let $v=xy$ and remember we are assuming $\mu$ can be rewritten as a function of $v$. Thus, \[ \mu_y = \frac{\bd \mu(v)}{\bd y} = \frac{d \mu}{dv} \frac{\bd v}{\bd y} = \frac{d\mu}{dv} \, \cdot x = x \mu', \] and \[ \mu_x = \frac{\bd \mu(v)}{\bd x} = \frac{d \mu}{d v} \frac{\bd v}{\bd y} = \frac{d\mu}{dv} \, \cdot y = y \mu', \] where $\mu'$ means the derivative with respect to $v$. Now $(*)$ becomes \[ x\mu'M + \mu M_y = y\mu'N + \mu N_x, \] which gives \[ \frac{\mu'}{\mu} = \frac{N_x - M_y}{xM-yN}. \] If the right hand side depends only on $v=xy$ then the assumption we are making is valid, and thus \[ \mu = e^{\int \frac{N_x - M_y}{xM-yN} \, dv}. \] Perhaps an example would help. \vspace{.25in} {\bf Example 7.} Solve \[ 5x^3 + \frac{1}{x}\cos xy + \frac{x^4 + \cos xy}{y} \frac{dy}{dx} = 0, \] with $y(1)=\pi$. \begin{proof}[Solution] Let $M = 5x^3 + \frac{1}{x}\cos xy$ and $N=\frac{x^4 + \cos xy}{y}$. Then \[ M_y = -\sin xy \hspace{0.5in}\&\hspace{0.5in} N_x = \frac{4x^3 - y \sin xy}{y} \] Thus, the given equation is not exact. We now search for an integration factor. \[ \mbox{Case 1.} \:\: \frac{M_y-N_x}{N} = \frac{-\frac{4x^3}{y}}{\frac{x^4+\cos xy}{y}} = \frac{-4x^3}{x^4 + \cos xy} \:\:\:\mbox{No good!} \] \[ \mbox{Case 2.} \:\: \frac{N_x-M_y}{M} = \frac{\frac{4x^3}{y}}{\frac{5x^4+\cos xy}{x}} = \frac{4x^4}{5x^4y + y \cos xy} \:\:\:\mbox{Rats!!} \] \[ \mbox{Case 3.} \:\: \frac{N_x-M_y}{xM-yN} = \frac{\frac{4x^3}{y}}{5x^4 + \cos xy -(x^4 + \cos xy)} = \frac{1}{xy}! \:\:\:\mbox{Eureka!!!} \] Let $v=xy$. Now, \[ \mu(v) = e^{\int \frac{1}{v} \, dv} = e^{\ln |v|+C} = C|v| = C|xy|; \] we will use $\mu = xy$ On ward! We multiply the original equation by $xy$ to get \[ 5x^4y + y \cos xy + (x^5 + x \cos xy)y' = 0. \] Let $M= 5x^4y + y \cos xy$ and $N=x^5 + x \cos xy$. We double check that it is in fact exact. \[ M_y = 5x^4 + \cos xy - xy \sin xy = N_x. \] Now the hunt is on for $\psi$! \[ \psi = \int M \, dx = x^5y + \sin xy + C_1(y) \] and \[ \psi = \int N \, dy = x^5y + \sin xy + C_2(x). \] Thus, $\psi = x^5y + \sin xy$ and the general solution is $x^5y + \sin xy =C$. Since $y(1) = \pi$ you can check that $C=\pi$. Thus, the solution is \[ x^5y + \sin xy = \pi. \] \end{proof} \vspace{.2in} \begin{center} {\sc Appendix A: More On Example 2.}\\{\sc [Optional Reading]} \end{center} Recall that Example 2 was not exact, but that it was noted that it is homogeneous. Here we will solve it and study the solution. \[ y' = \frac{-2x-2y}{x+2y} =\frac{-2-2y/x}{1+2y/x} = \frac{-2-2v}{1+2v},\] where v = y/x. It follows that $y' = v + xv'$. Now we have \[ x\frac{dv}{dx} = \frac{-2-2v}{1+2v} - v =\frac{-2-2v}{1+2v} - v \frac{1+2v}{1+2v} = \frac{-2-3v-2v^2}{1+2v}. \] Thus, \[ -\int\frac{1+2v}{2+3v+2v^2} \, dv = \int\frac{1}{x}\,dx. \] We rewrite the left integrand as \[ \frac{1}{2} \frac{4v+3-1}{2v^2+3v+2} = \frac{1}{2}\left(\frac{4v+3}{2v^2+3v+2}-\frac{1}{2v^2+3v+2} \right). \] Now, \[ \int \frac{4v+3}{2v^2+3v+2}\, dv = \ln|2v^2+3v+2| + C, \] and \[ \int \frac{1}{2v^2+3v+2}\, dv =\int \frac{1}{\left(\sqrt{2}v+\frac{3}{2\sqrt{2}}\right)^2 +7/8} \, dv. \] Let $u=\sqrt{2}v+\frac{3}{2\sqrt{2}}$. Then $du =\sqrt{2} dv$. The last integral becomes \[ \frac{1}{\sqrt{2}} \int \frac{du}{u^2+7/8} = \frac{4\sqrt{2}}{7} \int\frac{du}{8u^2/7 + 1}. \] Next let $w=\frac{2\sqrt{2}u}{\sqrt{7}}$. Then $dw = \frac{2\sqrt{2}du}{\sqrt{7}}$. Thus, the last integral becomes \[ \frac{2}{\sqrt{7}} \int \frac{dw}{w^2+1} = \frac{2}{\sqrt{7}} \arctan (w) + C. \] Putting all this together gives \[ \ln|x| + C = -\frac{1}{2} \ln|2v^2+3v+2| +\frac{1}{\sqrt{7}}\arctan\left(\frac{4v+3}{\sqrt{7}}\right). \] Now we find $C$, Recall we were given $y(1)=1$. Since $v=y/x$ we get $v=1$. Thus, \[ 0+C = \ln \left( \frac{1}{\sqrt{7}}\right) + \frac{1}{\sqrt{7}}\arctan \left(\sqrt{7}\right). \] Thus, our solution is given by the relation \[ \ln x = -\frac{1}{2} \ln (2v^2+3v+2) + \frac{1}{\sqrt{7}} \arctan\left( \frac{4v+3}{\sqrt{7}}\right) \]\[-\ln \left(\frac{1}{\sqrt{7}}\right) - \frac{1}{\sqrt{7}} \arctan\left( \sqrt{7}\right), \] where $v=y/x$. \vspace{.2in} I tried to plot this, and some simplifications, using {\tt implicitplot} in Maple 17 and using WolframAlpha online. Neither could do anything with it. So, I punted and used Maple to numerically plot the solution curve. Here is the command and the plot it produced. \vspace{.2in} $>${\tt DEplot(diff(y(x),x)=(-2*x-2*y(x))/(x+2*y(x)),y(x),}\\{\tt x=1.0..2.0,[[y(1)=1]],y=-2.0..1.0,arrows=line,color=black,}\\{\tt linecolor=red);} \vspace{.2in} \begin{center}\vspace*{-0.5in} \includegraphics[height=7in]{Figs/ExactEx2Plot.pdf} \end{center} \vspace*{-3in} It also produced this warning: ``Warning, plot may be incomplete, the following errors(s) were issued: cannot evaluate the solution further right of 1.9201778, probably a singularity''. If you look at the original differential equation you may notice that $y'(x)$ is undefined when $x=-2y$. Study the graph. At $x$ equal about 2, $y$ is just about $-1$. Our solution curve ends and is not valid passed this point. }\end{document}