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Homework Help: First-Order Linear Differential Equation Problem

  1. Sep 11, 2005 #1
    Consider the equation [tex] Ly'+Ry=E\sin(\omega x) [/tex] , where [tex] L,R,E,\omega [/tex] are positive constants.
    1) Compute the solution [tex] \phi [/tex] satisfying [tex] \phi(0)=0 .[/tex]
    2) Show that this solution may be written in the form [tex] \phi(x)=\frac{E\omega L}{R^2+\omega^2L^2}e^{-Rx/L}+\frac{E}{\sqrt{R^2+\omega^2L^2}}\sin(\omega x-\alpha) ,[/tex]
    where [tex] \alpha [/tex] is the angle satisfying [tex] \cos (\alpha)=\frac{R}{\sqrt{R^2+\omega^2L^2}} [/tex] , [tex] \sin(\alpha)=\frac{\omega L}{\sqrt{R^2+\omega^2L^2}}. [/tex]

    I have already solved part (1) of this problem but am having some difficulty converting the solution that I computed for part (1) to that which will allow me to answer part (2). A summary of the work that I did to solve part (1) can be found immediately following this sentence.

    [tex] Ly'+Ry=E\sin(\omega x) \iff y'=\frac{Ry}{L}=\frac{E\sin(\omega x)}{L} [/tex] .
    [tex] Let \ a(x)=R/L \implies \int a(x)dx=Rx/L .[/tex] [tex] \therefore e^{\int a(x)dx}=e^{Rx/L}. [/tex]

    [tex] e^{Rx/L}\left[y'+\frac{Rx}{L}\right]=e^{Rx/L}\left[\frac{E\sin(\omega x)}{L}\right] \implies \left(e^{Rx/L}y\right)'=\frac{E}{L}\left[e^{Rx/L}\sin(\omega x)\right] [/tex]
    [tex] \implies \int \left(e^{Rx/L}y\right)'dx=\frac{E}{L}\int e^{Rx/L}\sin(\omega x)dx=\frac{E}{L}\left[\frac{1}{(R/L)^2+\omega^2}\right]\left[e^{Rx/L}\right]\left[\frac{R}{L}\sin(\omega x)-\omega\cos(\omega x)\right]+C [/tex]
    [tex] =E\left[\frac{R\sin(\omega x)-L\omega\cos(\omega x)}{R^2+\omega^2L^2}\right]e^{Rx/L}+C \implies y=\phi(x)=E\left[\frac{R\sin(\omega x)-L\omega\cos(\omega x)}{R^2+\omega^2L^2}\right]+Ce^{-Rx/L}. [/tex]

    [tex] \phi(0)=E\left[\frac{0-L\omega}{R^2+\omega^2L^2}\right]+Ce^0=-E\left[\frac{0-L\omega}{R^2+\omega^2L^2}\right]+C=0 \implies C=E\left[\frac{0-L\omega}{R^2+\omega^2L^2}\right] .[/tex]

    [tex] \therefore \phi(x)=E\left[\frac{R\sin(\omega x)-L\omega\cos(\omega x)}{R^2+\omega^2L^2}\right]+E\left[\frac{L\omega}{R^2+\omega^2L^2}\right]e^{-Rx/L}. [/tex]

    What I cannot seem to figure out how it is that I am supposed to convert the first component of the solution, [tex] E\left[\frac{R\sin(\omega x)-L\omega\cos(\omega x)}{R^2+\omega^2L^2}\right] [/tex] , into [tex] \frac{E}{\sqrt{R^2+\omega^2L^2}}\sin(\omega x-\alpha) [/tex] as is written in part (2) of the above problem. It would be appreciated if someone could provide me with some additional insight into how to complete part (2) (it appears that I am probably not seeing what it is that I need to do in order to solve the remainder of the problem). Prompt responses are greately appreciated. Thank you for your time.
    Last edited: Sep 11, 2005
  2. jcsd
  3. Sep 11, 2005 #2

    James R

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    Note that

    [tex]\sin(\omega x - \alpha) = \sin(\omega x)\cos(\alpha) - \cos(\omega x)\sin(\alpha)[/tex]

    Does that help?
  4. Sep 12, 2005 #3
    Thank you for posting that identity. I have made some modifications to the original post-it appears that some typos were made while I was in the process of typing. All of the equations should now look more hospitable. Thank you for your time.
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