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Homework Help: 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


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    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


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


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    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


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    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


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    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


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    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.

  15. Jan 31, 2014 #14


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    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


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    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}##.
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