My intent is to create a thread for people interested in Differential Equations. However, I will explicitly state that I am only a student of this class myself and that many things could end up being incorrect or an improper way to present the material. I will merely be going along with the class, mostly using excerpts and questions from the book, "Elementary Differential Equations and Boundary Value Problems: Seventh Edition," by William E. Boyce and Richard C. DiPrima. So truthfully, this is more for myself. Looking things up and explaining it to others seems to be the best way to learn. If people have any questions or comments, feel free to share. Also, I know there are many knowledgeable people on this board, so be sure to correct me or make suggestions. This will require knowledge of Calculus but don't be shy to ask if there is something that you are unsure of. First, a little background; What is a Differential Equation? A Differential Equation is simply an equation containing a derivative. Classifications: Ordinary Differential Equations (ODE) - Equations that appear with ordinary derivatives (single independent variable, could have multiple dependent variables). Examples: [tex] \frac {dy} {dt} = ay - b [/tex] [tex] a \frac {dy_1} {dx} + b \frac {dy_2} {dx} + cy_1 = dy_2 = e [/tex] Partial Differential Equations (PDE) - Equations that appear with partial derivatives (multiple independent variable). Examples: [tex] \alpha^2 [ \frac {\partial^2 u(x,t)} {\partial x^2} ] = \frac {\partial u(x,t)} {\partial t} [/tex] [tex] \frac {\partial^2 V(x,y)} {\partial x^2} + \frac {\partial^2 V(x,y)} {\partial y^2} = 0 [/tex] Don't let any of this frighten you. Math is always scary when looked at a glance with a bunch of undefined variables. Linear and Nonlinear The ordinary differential equation: [tex] F(t, y, y', ..., y^{(n)}) = 0 [/tex] is said to be linear if F is a linear function of the variables y, y',..., y^{n} (Dependant variable must be first order). Thus the general linear ordinary differential equation of order n is: [tex] a_0 (t) y^{(n)} + a_1 (t) y^{(n-1)} + ... + a_n (t) y = g(t) [/tex] where ^{(n)} is not the power of but the nth derivative. An example of a simple Nonlinear ODE would simply be: [tex] y \frac {dy} {dx} = x^4 [/tex] This concludes the introduction. I may or may not write the next chapter tonight. However, a question, does anyone know an easier way for writing math on the computer and one that looks less confusing. I know I will have difficulty finding some things, especially subscripts and superscripts. Anyone know a better way to denote these?
Sounds great! Tutorials like this have been very successful here. Howto make math symbols: https://www.physicsforums.com/announcement.php?forumid=73 You can make subscripts and subscripts by using these tags [ sup ] content [ /sup ] [ sub ] content [ /sub ] * no spaces
First Order Differential Equations "This chapter deals with differential equations of the first order [tex] \frac {dy} {dt} = f(t,y) [/tex] where f is a given function of two variables. Any differentiable function y = Φ(t) that satisfies this equation for all t in some interval is called a solution." Linear Equations with Variable Coefficients Using the previous example for ODE (dy/dx = ay + b) and replacing the constants we write the more general form: [tex] \frac {dy} {dt} + p(t) y = g(t) [/tex] or [tex] y' + p(t) y + g(t) = 0 [/tex] where p(t) and g(t) are given functions of the independent variable t. Special cases: If p(t) = 0 then, [tex] y' = g(t) [/tex] and the integral is easily taken; [tex] \frac {dy} {dt} = g(t) [/tex] [tex] \int (\frac {dy} {dt}) dt = \int g(t) dt [/tex] [tex] y = \int g(t) dt + C [/tex] If g(t) = 0, then, [tex] y' = p(t) y [/tex] and the integral is once more relatively easy to take; [tex] \frac {dy} {dt}= p(t) y [/tex] [tex] \int \frac {dy} {y} = \int p(t) dt [/tex] [tex] ln|y| = \int [p(t) dt] + C [/tex] [tex] e^{ln|y|} = e^{\int [p(t) dt] + C} [/tex] [tex] y = Ke^{\int [p(t) dt] + C}, K = ± e^{C} [/tex] However, if neither p(t) or g(t) are zero in the general equation, a function µ(t) (the integrating factor) is used to solve the equation; [tex] \mu (t) \frac {dy} {dt} + \mu (t) p(t) y = \mu (t) g(t) [/tex] where now the left hand side will be a known derivative [tex] \mu (t) \frac {dy} {dt} + \mu (t) p(t) y = \frac {d} {dt}[\mu (t)y] [/tex] so that, in theory, you end up with; [tex] \frac {d} {dt}[\mu (t)y] = \mu (t) g(t) [/tex] Since µ(t) must be carefully chosen to make the previous statement true, let us find it. [tex] \frac {d} {dt}[\mu (t)y] = \mu (t) \frac {dy} {dt} + \mu (t)p(t)y [/tex] [tex] \frac {d} {dt}[\mu (t)y] = \mu (t) \frac {dy} {dt} + \frac {d \mu (t)} {dt} y [/tex] Where the latter is simply the derivative of µ(t)y in general form. [tex] \mu (t) \frac {dy} {dt} + \mu (t) p(t) y = \mu (t) \frac {dy} {dt} + \frac {d\mu (t)} {dt} y [/tex] Subtracting µ(t)(dy/dt) from both sides [tex] \frac {d \mu (t)} {dt} y = \mu (t) p(t) y [/tex] Cancel y [tex] \frac {d \mu (t)} {dt} = \mu (t) p(t) [/tex] [tex] \frac {d \mu (t)} {\mu (t)} = p(t) dt [/tex] [tex] \int \frac {d \mu (t)} {\mu (t)} = \int p(t)d t [/tex] [tex] ln|\mu (t)| = \int p(t) dt [/tex] The constant C is arbitrary and can be dropped to form the equation, µ(t) = e^{∫[p(t)dt]} So the integrating factor µ(t) can always be found by the last equation. Lets try some problems together. Ex1. y' + 2y = 3 In this equation p(t) = 2 and g(t) = 3. Since neither of them are zero, use an the integrating factor µ(t) to create a differentiable equation, µ(t)y' + µ(t)2y = µ(t)3 Solve µ(t)to be, µ(t) = e^{∫[p(t)dt]} µ(t) = e^{∫[2dt]} µ(t) = e^{2t} Plug the value of µ(t) back into the equation to obtain, e^{2t}y' + e^{2t}2y = e^{2t}3 Recognize that the left hand side of the equation is merely [ye^{2t}]', [ye^{2t}]' = d/dt[ye^{2t}] = 3e^{2t} ∫d/dt[ye^{2t}] = ∫3e^{2t} ye^{2t} = (3/2)e^{2t} + C y = (3/2) + Ce^{-2t} Ex2. y' + (1/2)y = 2 + t µ(t)y' + (1/2)µ(t)y = (2 + t)µ(t) µ(t) = e^{∫[p(t)dt]} = e^{t/2} e^{t/2}y' + (1/2)e^{t/2}y = 2e^{t/2} + te^{t/2} d/dt[e^{t/2}y] = 2e^{t/2} + te^{t/2} e^{t/2}y = ∫[2e^{t/2} + te^{t/2}]dt e^{t/2}y = 4e^{t/2} + ∫[te^{t/2}]dt Using integration by parts, u = t, du = dt v = 2e^{t/2}, dv = e^{t/2}dt ∫[te^{t/2}]dt = 2te^{t/2} - ∫[2e^{t/2}]dt ∫[te^{t/2}]dt = 2te^{t/2} - 4e^{t/2} + C e^{t/2}y = 4e^{t/2} + 2te^{t/2} - 4e^{t/2} = 2te^{t/2} + C y = 2t + Ce^{-t/2} For initial value problems it is easy to solve for C. Taking the last problem, solve if y(0) = 2 2 = 2(0) + Ce^{-(0)/2} = C C = 2 Therefore, y = 2t + 2e^{-t/2} That is enough for now. Here are some problems to practice on if you so wish. 1.) y' + 3y = t + e^{-2t} 2.) 2y' + y = 3t, hint: rewrite to fit general equation y' + p(t)y = g(t) 3.)t^{3}(dy/dt) + 4t^{2}y = e^{-t}, y(-1) = 0
Well I had to dig out my copy of Boyce and Diprima (2nd Edition!) to follow your development, it all works out as you have presented. I will follow along with you, relearning what I have not see for a number of years, and perhaps able to help you out if you hit some rough spots.
Thank you for your participation. Your solution is infact correct except for the constant is missing. No biggy, I always forget those too. Did you find it hard to follow without the book and should I have presented this more clearly some how? I'm glad to know that someone else knows this stuff. I have some trouble understanding finding the interval for nonlinear functions for which a solution exists, so if I havn't figured it out by the time I do the write up, perhaps you can help.
It was very sloppy of me to leave off the constant, sorry about that. I was a bit confused by your presentation as it is light on connective text. Where you presented [tex] \frac {d} {dt}[\mu (t)y] = \mu (t) g(t)[/tex] I was thrown for a bit. My copy of B&D helped out. The fact is everything you wrote is absolutely correct. I have taken grad level ODE & PDE courses in the dim and distant past ('86-'88 time frame) So should be able to dredge up some long buried knowledge to help out. I have always found Differential Equations to be interesting, you might say they are where math and reality meet. With a good back ground in Diff Eqs and some numerical methods you can do dang near anything. edit: corrected symbols
Yes, your answer is correct. It actually took me a while to get it. Out of curiousity, why did you change to using the variable s? Are you just used to using it and forgot that it was in terms of t or can this be done?
In the integral [tex] \int \mu (s) g(s)ds [/tex] The variable, s, is what is called a dummy variable, it can be anything. You will see this frequenty.
Integral, for the last part of your solution to the thrid exercise, I get y(t)=t^{-4} [inte]te^{-t}dt=t^{-4}[-te^{-t}-e^{-t} + C] So the t^{-4} is distributed through the C. y(t)=-t^{-4}e^{-t}(t + 1)+ t^{-4}C Applying the initial condition, I get C=0, so y(t)=-t^{-4}e^{-t}(t + 1). We get the same result, but through different ways.
Separable Equations I'm terribly sorry, I have been entirely too busy recently but I'm back, for the moment. So now that we know how to differentiate a first order linear equation with variable coefficients, let us move on to linear separable equations. In the last section we used the form dy/dt = ay + b, where the more general form of first order equation is; dy/dx = f(x,y) If this function can be rewritten in the form M(x) + N(y)dy/dx = 0 where M is a function of x only and N is a function of y only, then the function is said to be separable. It can also be written in the form of M(x)dx + N(y)dy = 0 It is as simple as that. Let's try a couple examples. Ex. 1 dy/dx = x^{2}/(1 - y^{2}) -x^{2} + (1 - y^{2})dy/dx = 0 ∫-x^{2}dx + ∫[(1 - y^{2})(dy/dx)]dx = ∫0dx -x^{3}/3 + (y - (y^{3})/3) = C Where the answers can be left in a few different forms; -x^{3}/3 + (y - (y^{3})/3) = C -x^{3} + 3y - y^{3} = 3C = C_{1} 3y - y^{3} = x^{3} + C_{1} The "cheating" way would be to cross multiply the initial equation. Although, this is not correct, it will end in the same answer. dy/dx = x^{2}/(1 - y^{2}) ∫(1 - y^{2})dy = ∫x^{2}dx y - (y^{3})/3 = x^{3}/3 + C 3y - (y^{3}) = x^{3} + C_{1} Let's try one with an initial condition Ex. 2 dy/dt = ycost/(1 + 2y^{2}), y(0) = 1 (1 + 2y^{2})dy/dt = ycost -cost + ((1 + 2y^{2})/y)dy/dt = 0 ∫-cos(t)dt + ∫[((1 + 2y^{2})/y)dy/dt]dt = ∫0dx -sint + ∫[y^{-1} + 2y]dy = C -sint + ln|y| + y^{2} = C ln|y| + y^{2} = sint + C ln|1| + 1^{2} = sin(0) + C C = 1 ln|y| y^{2} = sint + 1 Again, this differential equation can be solved the "cheating" way by cross multiplying, dy/dt = ycost/(1 + 2y^{2}), y(0) = 1 [(1 + 2y^{2})/y]dy = cos(t)dt ∫(y^{-1} + 2y)dy = ∫cos(t)dt ln|y| y^{2} = sint + C If you so desire, here are some problems 1. y' = x^{2}/y 2. xdx + ye^{-x}dy = 0, y(0) = 1 3. y^{2}(1 + x^{2})^{1/2} + arcsinx dx
The "cheating" way would be to cross multiply the initial equation. Although, this is not correct, it will end in the same answer. ____________ I've often wondered why this is the case. dy and dx are variables right? So why can't they be treated as such?
Here's my solution to the first one: dy/dx=x^{2}/y y(dy/dx)=x^{2} [inte] y(dy/dx)dx = [inte] x^{2}dx (1/2)y^{2}= (1/3)x^{3} + C_{1} y^{2}= (2/3)x^{3} + C_{2}
"I've often wondered why this is the case. dy and dx are variables right? So why can't they be treated as such?" x is a variable, y is a funtion of x. By cross multiplying you would be treating y like a variable. And your answer is correct.
Interval of Solution for Nonlinear First Order Equations I will mention right now, I have some difficulty understanding this part so I will have even more difficulty explaining it and will probably need a little help. The good news, this isn't the most important topic in my opinion. The reason this is applicable is that it is often easier to find the existence and uniquness of a solution without having to work out the actual problem. Theorm: Let the function f and ∂f/∂y be continuous in some rectangle [alpha] < t < [beta], Γ < y < δ containing the point (t_{0}, y_{0}). Then, in some interval t_{0} - h < t < t_{0} + h contained in [alpha] < t < [beta], there is a unique solution y = φ(t) of the initial value problem y' = f(t,y), y(t_{0}) = y_{0} Assuming that both the function and it's partial deivative with respect to y are continuous in a rectangle R; |t - t_{0}| < A, |y - y_{0}| < B; Let: M = max |f(t,y)|, (t,y) is in R C = min (A,B/M) Then: There is one and only one solution y(t) valid for t_{0}- C < x < t_{0}+ C Ex. y' = y^{2}, y(0) = 1 f(t,y) = y^{2} ∂f/∂y = 2y M = max |f(t,y)|, (t,y) is in R C = min (A,B/M) Since A can be made infinately large find the max for B M = max|(1 + B)^{2}| C = B/M C = B/(1 + B)^{2} C = 1/4 If you have no idea where the answer came from, you are in the same boat as I. If you do know, please share. Perhaps another example Ex. 2 y' = (t^{2} +y^{2})^{3/2}, y(t_{0}) = y_{0} ∂f/∂y = 3y(t^{2} + y^{2})^{1/2} M = max|(t^{2} +y^{2})^{3/2}| M = [(t_{0} + A)^{2} + (y_{0} + B)^{2}]^{3/2} C = min(A,B/M) C = min(A, B/[(t_{0} + A)^{2} + (y_{0} + B)^{2}]^{3/2}) Also, does anyone know how to display determinants or matracies on the computer?