1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

First order differential equations

  1. Jan 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider the first order differential equation

    [itex]\frac{dx(t)}{dt} + ax(t) = f(t), x(0) = x_{0}, t\geq0[/itex] ​

    Suppose the "input signal" [itex] f(t)=e^{-t}, t\geq0 [/itex]. (a) Find the solution to the equation. Find a condition on the parameter a so that the solution of the (forced) system approaches zero as t→∞.

    2. Relevant equations
    [itex]\frac{dy}{dt} + p(x)y = 0 [/itex]


    3. The attempt at a solution

    Setup as a homogenous equation therefore f(t) = 0.
    [itex]\frac{dx(t)}{dt} + ax(t) = 0 [/itex]

    [itex]\frac{dx(t)}{x(t)} = -a*dt [/itex]

    [itex] ln(x(t)) = -at [/itex]

    [itex] x(t) = e^{-at} [/itex]

    I don't know how to proceed any further...
     
  2. jcsd
  3. Jan 30, 2014 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What you found satisfies the homogeneous equation. You are still missing an integration constant.
    Now you need one particular solution that satisfies the inhomeogeneous equation.
    A times homogeneous solution + particular solution is ' the ' general solution
     
  4. Jan 30, 2014 #3
    Is the integration constant [itex] e^{\int x(t) dt} [/itex] or [itex] e^{\int a dt} [/itex]?

    I guess the only integration factor that makes sense is the later of the two.
     
    Last edited: Jan 30, 2014
  5. Jan 30, 2014 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, I mean [itex] x(t) = Ce^{-at} [/itex] also satisfies the homogeneous equation.
     
  6. Jan 30, 2014 #5

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Next thing to try: [itex] C = C(t) [/itex]. Put this in the inhomogeneous equation and see what equation you get for C.
     
  7. Jan 30, 2014 #6
    I'm sorry but you've now lost me. Why am I trying C=C(t)?
     
  8. Jan 30, 2014 #7

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Well, the [itex] e^{-at} [/itex] part of [itex] x(t) = C(t)e^{-at} [/itex] helps you get rid of the [itex] +a x(t) [/itex] term. In return, you are left with a differential equation for [itex] C(t) [/itex] that you may be able to solve.
     
  9. Jan 30, 2014 #8

    pasmith

    User Avatar
    Homework Helper

    Either use Method of undetermined coefficients or use the integrating factor [itex]e^{at}[/itex].
     
  10. Jan 30, 2014 #9
  11. Jan 31, 2014 #10
    How can you tell when a first differential needs to use an integrating factor?
     
  12. Jan 31, 2014 #11
    The differential equation you have presented is a linear differential equation of first order and this immediately tells us to use the integrating factor.
     
  13. Jan 31, 2014 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    In this case, you don't need an integrating factor. Just put ##x(t)=Ce^{-t}## into the equation and solve for C. You can't always do that but it's a nice shortcut here. Can you see why you could guess it would work? In general, guessing a trial solution is often a good option. When you can.
     
  14. Jan 31, 2014 #13
    Look at the left side of your equation. "a" is a constant, so it is certainly going to integrate nicely. If "a" can be integrated easily with respect to t, that's your first indicator. Now look at the right side of the equation. It's a function only of t. So, when you multiply both sides by the integrating factor, the right side will still only be a function only of t (that can be integrated). That's your second indicator.

    Chet
     
  15. Jan 31, 2014 #14

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Perhaps it's good to give a general derivation of the integrating factor for 1st-order linear ODEs. The standard form of such an ODE is
    [tex]y'(x)+p(x) y(x)=q(x),[/tex]
    where [itex]p[/itex] and [itex]q[/itex] are given functions and the equation is to besolved for [itex]y(x)[/itex]. We look for the general solution.

    The equation would be much easier, if we had a total derivative on the left-hand side, but that's not too difficult to achieve. You just substitute
    [tex]z(x)=v(x) y(x).[/tex]
    From the product rule you get
    [tex]z'(x)=v(x) y'(x)+v'(x) y(x).[/tex]
    Multiplying the ODE with [itex]v(x)[/itex] gives
    [tex]v(x) y'(x)+p(x) v(x) y(x)=q(x) v(x).[/tex]
    To make the left-hand side equal to [itex]z'(x)[/itex] you obviously have to determine [itex]v(x)[/itex] so that
    [tex]v'(x)=p(x) v(x)[/tex]
    This is very easily solved, because you can write
    [tex]\frac{\mathrm{d}}{\mathrm{d} x} \ln \left (\frac{v(x)}{v_0} \right )=p(x).[/tex]
    This gives
    [tex]v(x)=v_0 \exp \left (\int_{0}^{x} \mathrm{d} x' p(x') \right ).[/tex]
    Here [itex]v_0 \neq 0[/itex] is an arbitrary integration constant.

    Thus, choosing [itex]v(x)[/itex] in this way we can write our ODE as
    [tex]z'(x)=q(x) v(x)[/tex]
    with the general solution
    [tex]z(x)=z_0 + \int_0^{x} \mathrm{d} x' q(x') v(x').[/tex]
    Finally we have
    [tex]y(x)=\frac{z(x)}{v(x)}=\frac{z_0}{v(x)}+\frac{1}{v(x)} \int_0^{x} \mathrm{d} x' q(x') v(x').[/tex]
    As you see, the choice of [itex]v_0[/itex] doesn't affect your solution in any way, because it only appears as an overall factor in front of the integration constant [itex]z_0[/itex], and this can be lumped into the overall integration constant of the general solution of the homogeneous equation. Finally we have the general solution of the ODE
    [tex]y(x)=\exp \left (-\int_0^{x} \mathrm{d} x' p(x') \right ) \left [y_0+ \int_0^x \mathrm{d} x' q(x') \exp \left ( \int_0^x \mathrm{d} x'' p(x'') \right ) \right].[/tex]
    Here [itex]y_0[/itex] is given by the initial-value condition [itex]y(0)=y_0[/itex].
     
  16. Feb 1, 2014 #15
    Thank you everyone! I finally figured it out using integrating factors. As for guessing, would you guess Ce^-t as a good solution because the function is forced by e^-t?
     
  17. Feb 1, 2014 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's a good guess because of the other side of the equation, if you differentiate ##e^{-t}## or multiply it by ##a##, you just get back multiples of ##e^{-t}##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: First order differential equations
Loading...