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Intro. to Differential Equations

  1. Sep 20, 2003 #1
    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.


    Ordinary Differential Equations (ODE) - Equations that appear with ordinary derivatives (single independent variable, could have multiple dependent variables).

    \frac {dy} {dt} = ay - b

    a \frac {dy_1} {dx} + b \frac {dy_2} {dx} + cy_1 = dy_2 = e

    Partial Differential Equations (PDE) - Equations that appear with partial derivatives (multiple independent variable).

    \alpha^2 [ \frac {\partial^2 u(x,t)} {\partial x^2} ] = \frac {\partial u(x,t)} {\partial t}

    \frac {\partial^2 V(x,y)} {\partial x^2} + \frac {\partial^2 V(x,y)} {\partial y^2} = 0

    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:
    F(t, y, y', ..., y^{(n)}) = 0
    is said to be linear if F is a linear function of the variables y, y',..., yn (Dependant variable must be first order). Thus the general linear ordinary differential equation of order n is:
    a_0 (t) y^{(n)} + a_1 (t) y^{(n-1)} + ... + a_n (t) y = g(t)

    where (n) is not the power of but the nth derivative.

    An example of a simple Nonlinear ODE would simply be:
    y \frac {dy} {dx} = x^4
    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?
    Last edited by a moderator: Sep 6, 2013
  2. jcsd
  3. Sep 21, 2003 #2
    Sounds great! Tutorials like this have been very successful here.

    Howto make math symbols:
    https://www.physicsforums.com/announcement.php?forumid=73 [Broken]

    You can make subscripts and subscripts by using these tags

    [ sup ] content [ /sup ]
    [ sub ] content [ /sub ]

    * no spaces
    Last edited: May 1, 2017
  4. Sep 21, 2003 #3
    First Order Differential Equations

    "This chapter deals with differential equations of the first order

    \frac {dy} {dt} = f(t,y)

    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:

    \frac {dy} {dt} + p(t) y = g(t)
    y' + p(t) y + g(t) = 0

    where p(t) and g(t) are given functions of the independent variable t.

    Special cases:
    If p(t) = 0 then,

    y' = g(t)

    and the integral is easily taken;

    \frac {dy} {dt} = g(t)

    \int (\frac {dy} {dt}) dt = \int g(t) dt

    y = \int g(t) dt + C

    If g(t) = 0, then,

    y' = p(t) y

    and the integral is once more relatively easy to take;

    \frac {dy} {dt}= p(t) y

    \int \frac {dy} {y} = \int p(t) dt

    ln|y| = \int [p(t) dt] + C

    e^{ln|y|} = e^{\int [p(t) dt] + C}

    y = Ke^{\int [p(t) dt] + C}, K = ± e^{C}

    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;

    \mu (t) \frac {dy} {dt} + \mu (t) p(t) y = \mu (t) g(t)

    where now the left hand side will be a known derivative

    \mu (t) \frac {dy} {dt} + \mu (t) p(t) y = \frac {d} {dt}[\mu (t)y]

    so that, in theory, you end up with;

    \frac {d} {dt}[\mu (t)y] = \mu (t) g(t)

    Since µ(t) must be carefully chosen to make the previous statement true, let us find it.

    \frac {d} {dt}[\mu (t)y] = \mu (t) \frac {dy} {dt} + \mu (t)p(t)y

    \frac {d} {dt}[\mu (t)y] = \mu (t) \frac {dy} {dt} + \frac {d \mu (t)} {dt} y

    Where the latter is simply the derivative of µ(t)y in general form.

    \mu (t) \frac {dy} {dt} + \mu (t) p(t) y = \mu (t) \frac {dy} {dt} + \frac {d\mu (t)} {dt} y

    Subtracting µ(t)(dy/dt) from both sides

    \frac {d \mu (t)} {dt} y = \mu (t) p(t) y

    Cancel y

    \frac {d \mu (t)} {dt} = \mu (t) p(t)

    \frac {d \mu (t)} {\mu (t)} = p(t) dt

    \int \frac {d \mu (t)} {\mu (t)} = \int p(t)d t

    ln|\mu (t)| = \int p(t) dt

    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.

    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) = e2t

    Plug the value of µ(t) back into the equation to obtain,

    e2ty' + e2t2y = e2t3

    Recognize that the left hand side of the equation is merely [ye2t]',

    [ye2t]' = d/dt[ye2t] = 3e2t
    ∫d/dt[ye2t] = ∫3e2t
    ye2t = (3/2)e2t + C
    y = (3/2) + Ce-2t

    y' + (1/2)y = 2 + t
    µ(t)y' + (1/2)µ(t)y = (2 + t)µ(t)

    µ(t) = e∫[p(t)dt] = et/2

    et/2y' + (1/2)et/2y = 2et/2 + tet/2
    d/dt[et/2y] = 2et/2 + tet/2
    et/2y = ∫[2et/2 + tet/2]dt
    et/2y = 4et/2 + ∫[tet/2]dt

    Using integration by parts,

    u = t, du = dt
    v = 2et/2, dv = et/2dt

    ∫[tet/2]dt = 2tet/2 - ∫[2et/2]dt
    ∫[tet/2]dt = 2tet/2 - 4et/2 + C

    et/2y = 4et/2 + 2tet/2 - 4et/2 = 2tet/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

    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.)t3(dy/dt) + 4t2y = e-t, y(-1) = 0
    Last edited: Jul 27, 2004
  5. Sep 21, 2003 #4
    Thanks Greg, I will run through it tomorrow and change it to make it more readable.
  6. Sep 21, 2003 #5


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    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.
  7. Sep 21, 2003 #6


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    Last edited by a moderator: Apr 20, 2017
  8. Sep 21, 2003 #7
    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.
  9. Sep 21, 2003 #8


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    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
    Last edited: Feb 23, 2007
  10. Sep 22, 2003 #9


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  11. Sep 22, 2003 #10
    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?
  12. Sep 22, 2003 #11


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    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.
    Last edited: Feb 23, 2007
  13. Sep 26, 2003 #12
    Integral, how come your solutions don't show up?
  14. Sep 26, 2003 #13


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    They are PDFs do you have acrobat reader installed?
  15. Sep 26, 2003 #14
    Yeah I do. I didn't realize your text was a link.
  16. Sep 27, 2003 #15
    Integral, for the last part of your solution to the thrid exercise, I get
    y(t)=t-4 [inte]te-tdt=t-4[-te-t-e-t + C]
    So the t-4 is distributed through the C.
    y(t)=-t-4e-t(t + 1)+ t-4C
    Applying the initial condition, I get C=0, so
    y(t)=-t-4e-t(t + 1). We get the same result, but through different ways.
    Last edited: Sep 27, 2003
  17. Sep 28, 2003 #16
    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 = x2/(1 - y2)

    -x2 + (1 - y2)dy/dx = 0
    ∫-x2dx + ∫[(1 - y2)(dy/dx)]dx = ∫0dx
    -x3/3 + (y - (y3)/3) = C

    Where the answers can be left in a few different forms;
    -x3/3 + (y - (y3)/3) = C
    -x3 + 3y - y3 = 3C = C1
    3y - y3 = x3 + C1

    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 = x2/(1 - y2)
    ∫(1 - y2)dy = ∫x2dx
    y - (y3)/3 = x3/3 + C
    3y - (y3) = x3 + C1

    Let's try one with an initial condition

    Ex. 2
    dy/dt = ycost/(1 + 2y2), y(0) = 1

    (1 + 2y2)dy/dt = ycost
    -cost + ((1 + 2y2)/y)dy/dt = 0
    ∫-cos(t)dt + ∫[((1 + 2y2)/y)dy/dt]dt = ∫0dx
    -sint + ∫[y-1 + 2y]dy = C
    -sint + ln|y| + y2 = C
    ln|y| + y2 = sint + C

    ln|1| + 12 = sin(0) + C
    C = 1

    ln|y| y2 = sint + 1

    Again, this differential equation can be solved the "cheating" way by cross multiplying,

    dy/dt = ycost/(1 + 2y2), y(0) = 1

    [(1 + 2y2)/y]dy = cos(t)dt
    ∫(y-1 + 2y)dy = ∫cos(t)dt
    ln|y| y2 = sint + C

    If you so desire, here are some problems

    1. y' = x2/y
    2. xdx + ye-xdy = 0, y(0) = 1
    3. y2(1 + x2)1/2 + arcsinx dx
    Last edited: Sep 28, 2003
  18. Sep 28, 2003 #17
    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?
  19. Sep 28, 2003 #18
    Here's my solution to the first one:
    [inte] y(dy/dx)dx = [inte] x2dx
    (1/2)y2= (1/3)x3 + C1
    y2= (2/3)x3 + C2
  20. Sep 28, 2003 #19
    "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.
  21. Sep 28, 2003 #20
    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.

    Let the function f and &part;f/&part;y be continuous in some rectangle [alpha] < t < [beta], &Gamma; < y < &delta; containing the point (t0, y0). Then, in some interval t0 - h < t < t0 + h contained in [alpha] < t < [beta], there is a unique solution y = &phi;(t) of the initial value problem
    y' = f(t,y), y(t0) = y0

    Assuming that both the function and it's partial deivative with respect to y are continuous in a rectangle R; |t - t0| < A, |y - y0| < B;

    M = max |f(t,y)|, (t,y) is in R
    C = min (A,B/M)

    There is one and only one solution y(t) valid for
    t0- C < x < t0+ C

    http://www.angelfire.com/tx5/extravagantdreams/Nonlinear_Interval.jpg [Broken]

    y' = y2, y(0) = 1
    f(t,y) = y2
    &part;f/&part;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' = (t2 +y2)3/2, y(t0) = y0
    &part;f/&part;y = 3y(t2 + y2)1/2

    M = max|(t2 +y2)3/2|
    M = [(t0 + A)2 + (y0 + B)2]3/2

    C = min(A,B/M)
    C = min(A, B/[(t0 + A)2 + (y0 + B)2]3/2)

    Also, does anyone know how to display determinants or matracies on the computer?
    Last edited by a moderator: May 1, 2017
  22. Sep 29, 2003 #21


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    Keep an eye on what you are doing in the above examples. You are finding a specific region upon which you can guarantee that a unique solution exists. This means you must provide some specific numbers to the bounds of the region.

    One thing that might help you a little, redraw your picture with the A and B intervals CENTERED on (t0,y0)

    In your worked example the interval for y is:
    B-y0<=y <= B+y0

    Since y0=1 we have:

    so if M= Max|f(t,y)| and f(t,y)= y2 The maximum value of f(t,y) on our interval is (B+1)2
    I believe that this is saying you are free to pick any value on the t axis for t0, once that value is defined, the value of f(t[0,y) will be fixed and you can guarantee the existence of a Max value for f(t,y)On an interval surrounding it.
    This follows from the definitions.
    To get this a value of B must be given or defined, since you are free to choose B to be anything, it looks like someone picked 1.

    Does this help?
    Last edited: Sep 29, 2003
  23. Sep 29, 2003 #22
    I guess I am getting confused because the book does not use
    M = max |f(t,y)|
    C = min (A,B/M)

    notation. From what I understand, all this really says is that you are trying to find the maximum interval of f(t,y), where the limiting factor is either A or B/M (which is denoted C). But why B/M, why not just B? This is what gets me. Is C a value in the horizontal direction? Does dividing B by M make it a horizontal value? And how do you choose B?

    I think I am making this too difficult. I fear it is one of those things where you just have to look at it.
  24. Sep 30, 2003 #23
    OD Exact Equations and Integrating Factors

    Let the functions M, N, My, Nx, where subscripts denote partial derivatives, be continuous in the rectangular region R: [alpha] < [beta], &Gamma; < y < &delta;. Then;

    M(x,y) + N(x,y)y' = 0

    is an exact differential equation in R if and only if

    My(x,y) = Nx(x,y)

    at each point of R. That is, there exists a function satisfying

    &part;[psi]/&part;x(x,y) = M(x,y), &part;[psi]/&part;y(x,y) = N(x,y)
    [psi]x(x,y) = M(x,y), [psi]y(x,y) = N(x,y)

    if and only if M and N satisfy

    My(x,y) = Nx(x,y)

    This means that there is a solution [psi](x,y) of the general equation M + Ny' = 0

    Proof of My(x,y) = Nx(x,y)
    We already defined

    [psi]x(x,y) = M(x,y), [psi]y(x,y) = N(x,y)

    we can compute the partial derivative of each to be

    [psi]xy(x,y) = My(x,y), [psi]yx(x,y) = Nx(x,y)

    Since My and Nx are continuous, it follows that [psi]xy and [psi]yx are continuous also, which also guarantees their equality.

    Finding [psi](x,y)
    when My(x,y) = Nx(x,y):
    Starting with the equations

    [psi]x(x,y) = M(x,y), [psi]y(x,y) = N(x,y)

    start with the first and integrate with respect to x

    [psi](x,y) = &int;M(x,y)dx + h(y)

    Where h is some function of y playing the role of an arbitrary constant. Now we must proove that h(y) can always be chosed so that [psi]y = N

    [psi]y(x,y) = &part;/&part;y[&int;M(x,y)dx + h(y)]
    [psi]y(x,y) = &int;My(x,y)dx + h'(y)

    Setting [psi]y = N we obtain

    N(x,y) = &int;My(x,y)dx + h'(y)

    Where we can then solve for h'(y)

    h'(y) = N(x,y) - &int;My(x,y)dx
    h(y) = Nx(x,y) - My(x,y)

    Then the general solution

    [psi](x,y) = &int;M(x,y)dx + &int;[N(x,y) - &int;My(x,y)dx]dy

    Ex. 1

    2xy3 + 3x2y2y' = 0

    Where M = 2xy3, N = 3x2y2
    Then My = 6xy2, Nx = 6xy2

    Since My = Nx
    This equations is exact and can be solved by the previous method.

    Start with

    [psi]x = M = 2xy3
    [psi] = &int;Mdx + h(y) = &int;2xy3dx + h(y)
    [psi] = x2y3 + h(y)

    Find h;
    Since we know that N = &part;[psi]/&part;y, differentiate both sides and substitude N

    N = &part;/&part;y[&int;Mdx + h(y)
    N = 3x2y2 + h'(y)

    h'(y) = 0, h(y) = C

    Plug h(y) back into the original [psi] equation
    [psi] = x2y3 + C

    [psi] = x2y3 = K

    Ex. 2
    Find the function [psi](x,y) of
    y' = -(ax + by)/(bx - cy)

    (bx - cy)y' = -(ax + by)
    (ax + by) + (bx - cy)y' = 0

    M = (ax + by), N = (bx - cy)
    My = b, Nx = b

    Since My = Nx, the equation is exact

    [psi]x = M = ax + by
    &int;[psi]xdx = &int;(ax + by)dx
    [psi] = (1/2)ax2 + byx + h(y)
    [psi]y = &part;/&part;y[(1/2)ax2 + byx + h(y)]
    [psi]y = bx + h'(y)

    [psi]y= N

    N = bx + h'(Y)

    (bx - cy) = bx + h'(y)

    h'(y) = -cy
    h(y) = -(1/2)cy2

    [psi] = (1/2)ax2 + byx - (1/2)cy2 = K

    Ex. 3
    (ycosx + 2xey) + (sinx + x2ey - 1)y' = 0

    M = ycosx + 2xey, N = sinx + x2ey - 1
    My = cosx + 2xey, Nx = cosx + 2xey

    My = Nx, so can be solved using the exact method

    Remember that
    M = [psi]x
    N = [psi]y

    &int;[psi]xdx = &int;(ycosx + 2xey)dx
    [psi] = ysinx + eyx2 + h(y)

    Taking a partial derivative of both sides with respect to y,

    [psi]y = sinx + x2ey + h'(y)

    [psi]y = N = sinx + x2ey - 1

    sinx + x2ey - 1 = sinx + x2ey + h'(y)
    h'(y) = -1
    &int;h'(y)dy = h(y) = -y

    [psi](x,y) = ysinx + eyx2 - y = k

    Integrating Factor
    If an equation is not exact it can be multyplied by an integrating factor [mu] so that it becomes exact.

    Starting with the general form
    M(x,y) + N(x,y)y' = 0

    My [x=] Nx

    There is an integrating factor [mu] such that
    [mu]M(x,y) + [mu]N(x,y)y' = 0

    ([mu]M)y = ([mu]N)x


    [mu]yM + [mu]My = [mu]xN + [mu]Nx
    M[mu]y - N[mu]x + (My - Nx)[mu] = 0

    We will not discuss finding [mu](x,y) since this is entirely too diffictuly for this course, so we shall stick to [mu] as a function of x or y only.

    If [mu] is a function of x;
    ([mu]M)y = [mu]My, ([mu]N)x = [mu]Nx + N(d[mu]/dx)

    Thus, if ([mu]M)yis to equal [mu]Nx;
    d[mu]\dx = (My - Nx)[mu]/N

    If the function (My - Nx)/N depends on x only then [mu] can also be a function of x only and an integrating factor has been found.

    The proceedure to finding [mu](y) is similar and the equation
    d[mu]\dx = -(My = Nx)[mu]/M
    is derived.

    Ex. 4
    (3xy + y2) + (x2 + xy)y' = 0
    M = 3xy + y2, N = x2 + xy
    My = 3x + 2y, Nx = 2x + y

    My [x=] Nx

    the equation is not seperable and an integrating factor must be found

    (My - Nx)/N dx = d[mu](x)/[mu](x)
    ln|[mu](x)| = &int;(3x + 2y - 2x - y)/(x2 + xy)dx
    ln|[mu](x)| = &int;(x + y)/x(x + y)dx
    ln|[mu](x)| = &int;(1/x)dx
    ln|[mu](x)| + ln|x|
    [mu](x) = x

    Since it is a function of x only, this can be used as the integrating factor. Now to solve the equations

    x(3xy + y2) + x(x2 + xy)y' = 0
    M = 3x2y + xy2, N = x3 + x2y
    My = 3x2 + 2xy, Nx = 3x2 + 2yx

    My = Nx
    So it is now exact.

    &part;/&part;y[&int;(3x2y + xy2)dx] = x3 + x2y
    &part;/&part;y[yx3 + (1/2)y2x2 + h(y)] = x3 + x2y
    x3 + x2y + h'(y) = x3 + x2y

    h'(y) = 0, h(y) = C

    yx3 + (1/2)y2x2 = K


    1. (2x + 3) + (2y - 2)y' = 0
    2. (9x2 + y - 1)dx - (4y - x)dy = 0, y(1) = 0
    3. (x2y = 2xy + y3)dx + (x2 + y2)dy = 0
    Last edited: Oct 2, 2003
  25. Sep 30, 2003 #24


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    This is covered pretty well in my copy of B&D.

    He is using Picard iterations to prove the existance of solutions. I'll cheat for this cause he says it better then I. Here is are links to scans of Boyce & Deprima 2nd Ed.http://home.comcast.net/~rossgr1/Math/B_D1.jpg
    Hope this helps, is this not in your book?
    Last edited by a moderator: Apr 20, 2017
  26. Oct 2, 2003 #25
    Second Order Linear Differential Equations

    A second order differential equation is linear if it is in the following form;
    y'' + p(t)y' + q(t)y = g(t)
    P(t)y'' + Q(t)y' + R(t)y = g(t)

    Where p, q, and g and functions of t

    If the problem has initial conditions it will be in the form of
    y(t) = y0, y'(t) = y'0.

    Let's start with the easiest:
    Homogeneous with Constant coefficients

    Homogeneous in this case means g(t) = 0,
    with constant coefficients mean P(t) = p, Q(t) = q, and R(t) = r
    Since makes the general equations;

    py'' + qy' + ry = 0

    Solving the homogeneous equation will later always provide a way to solve the corresponding nonhomogeneous problem.

    I'm not going to proove all this but you can take the kernal of this funtion as

    ar2 + br + c = 0

    and you can, so to speak, find the roots of this funtion.

    r1,2 = (-b &plusmn; &radic;(b2 -4ac))/2a

    r1 = (-b + &radic;(b2 -4ac))/2a
    r2 = (-b - &radic;(b2 -4ac))/2a

    Assuming that these roots are real and different then;
    y1(t) = er1t
    y2(t) = er2t

    and y = C1y1(t) + C2y2(t)

    Which comes from the initial deriviation. If someone really really wants to know, I will show you.

    y = C1er1t + C2er2t

    This is your general solution.
    C1 and C2 can be solved for if initial conditions y(t) and y'(t) are given in the following manner;

    y = C1er1t + C2er2t

    y(t) = y0

    y0 = C1er1t + C2er2t


    y' = r1C1er1t + r2C2er2t

    y'(t) = y'0

    It is also possible to varify your solution by using
    y = C1er1t + C2er2t
    y' = r1C1er1t + r2C2er2t
    y'' = r12C1er1t + r22C2er2t

    and pluging them back into the equation
    ay'' + by' + cy = 0

    Ex. 1
    y'' + 5y' +6y = 0, y(0) = 2, y'(0) = 3

    1r2 + 5r + 6 = 0

    (r + 3)(r + 2)

    r1 = -3
    r2 = -2

    y = C1e-3t + C2e-2t

    2 = C1e-3(0) + C2e-2(0)
    2 = C1 + C2

    y' = -3C1e-3t - 2C2e-2t

    3 = -3C1e-3(0) - 2C2e-2(0)
    3 = -3C1 - 2C2

    |+1 +1 +2|
    |- 3 - 2 +3| ~

    |+1 +1 +2|
    |+0 +1 +9| ~

    |+1 +0 - 7|
    |+0 +1 +9| ~

    C1 = -7
    C2 = 9

    y = -7e-3t + 9e-2t
    y' = 21e-3t - 18e-2t
    y'' = -63e-3t + 36e-2t

    y'' + 5y' +6y = 0
    (-63e-3t + 36e-2t) +5(21e-3t -18e-2t) =6(-7e-3t + 9e-2t)

    e-3t(105 - 63 - 42) + e-2t(36 - 18 - 18) = 0
    Last edited: Oct 2, 2003
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