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July 7th, 2011, 10:32 AM   #1
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Differential equation problem

[attachment=0:3arcjy5l]untitled_1.png[/attachment:3arcjy5l]

I have the following problem from a book I'm reading, it occurs several chapters before integration and DE's are covered!

The current in the RCL-circuit above must satisfy the equation

$L\frac{d^2I}{dt^2}\,+\,R\frac{dI}{dt}\,+\,\frac{I} {C}\,=\,\frac{dE}{dt}$

If $E\,=\,E_0\sin(wt)$ where $E_0$ and $w$ are constants, show that a solution is

$I\,=\,f(t)\,=\,e^{(-Rt/(2L))}$$A\cos(vt)\,+\,B\sin(vt)$$\,+\,\frac{E_0}{Z }\sin(wt\,-\,\phi)$

where $A$ and $B$ are any constants whatsoever and

$v\,=\,\sqrt{\frac{1}{LC}\,-\,\frac{R^2}{4L^2}},\,Z\,=\,\sqrt{R^2\,+\,$$wL\,-\,\frac{1}{wC}$$^2},\,\tan(\phi)\,=\,\frac{wL-1/(wC)}{R}$
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 July 7th, 2011, 01:53 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,208 Thanks: 516 Math Focus: Calculus/ODEs Re: Differential equation problem Perhaps they intend for you to take the given solution, find its first and second derivatives with respect to t, then substitute into the ODE to show that it satisfies the equation. However, here is how I would derive the solution from the given information: Given $E=E_0\sin$$\omega t$$$ then $\frac{dE}{dt}=E_0\omega\cos$$\omega t$$$ and the ODE is then: $L\frac{d^2I}{dt^2}+R\frac{dI}{dt}+\frac{I}{C}=E_0\ omega\cos$$\omega t$$$ First, we find the solution $f_h(t)$ to the corresponding homogeneous equation, whose associated auxiliary equation is: $Lr^2+Rr+\frac{1}{C}=0$ Application of the quadratic formula yields: $r=\frac{-R\pm\sqrt{R^2-\frac{4L}C{}}}{2L}=-\frac{R}{2L}\pm i\sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}$ With $v=\sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}$ we have $r=-\frac{R}{2L}\pm v\cdot i$ Since the auxiliary equation has complex conjugate roots, this means: $f_h(t)=e^{-\frac{R}{2L}t}$$A\cos(vt)+B\sin(vt)$$$ where A and B are constants. Now we may use the method of undetermined coefficients to find the particular solution $f_p(t)$. We begin by assuming it will be of the form: $f_p(t)=k_1\cos$$\omega t$$+k_2\sin$$\omega t$$$ where $k_i$ are the coefficients to be determined. Differentiating with respect to t, we find: $f_p'(t)=-k_1\omega\sin$$\omega t$$+k_2\omega\cos$$\omega t$$=\omega$$k_2\cos\(\omega t$$-k_1\sin$$\omega t$$\)$ $f_p''(t)=-k_1\omega^2\cos$$\omega t$$-k_2\omega^2\sin$$\omega t$$=-\omega^2f_p(t)$ Substituting into the original ODE we have: $-L\omega^2f_p(t)+R\omega$$k_2\cos\(\omega t$$-k_1\sin$$\omega t$$\)+\frac{f_p(t)}{C}=E_0\omega\cos$$\omega t$$$ $$$\frac{1}{C}-L\omega^2$$$$k_1\cos\(\omega t$$+k_2\sin$$\omega t$$\)+R\omega$$k_2\cos\(\omega t$$-k_1\sin$$\omega t$$\)=E_0\omega\cos$$\omega t$$$ $$$k_1\(\frac{1}{C}-L\omega^2$$+k_2R\omega\)\cos$$\omega t$$+$$k_2\(\frac{1}{C}-L\omega^2$$-k_1R\omega\)\sin$$\omega t$$=E_0\omega\cos$$\omega t$$$ Equating coefficients across the equation yields the system: $k_1$$\frac{1}{C}-L\omega^2$$+k_2R\omega=E_0\omega$ $k_2$$\frac{1}{C}-L\omega^2$$-k_1R\omega=0$ Solving the 2nd equation for $k_1$ we find: $k_1=\frac{k_2$$\frac{1}{C}-L\omega^2$$}{R\omega}$ Substituting for $k_1$ into the 1st equation gives: $\frac{k_2$$\frac{1}{C}-L\omega^2$$}{R\omega}$$\frac{1}{C}-L\omega^2$$+k_2R\omega=E_0\omega$ Solving for $k_2$ gives: $k_2=\frac{E_0R\omega^2}{$$\frac{1}{C}-L\omega^2$$^2+$$R\omega$$^2}=\frac{E_0R}{$$\frac{1 }{\omega C}-L\omega$$^2+R^2}$ And hence: $k_1=\frac{k_2$$\frac{1}{\omega C}-L\omega$$}{R}=\frac{E_0$$\frac{1}{\omega C}-L\omega$$}{$$\frac{1}{\omega C}-L\omega$$^2+R^2}$ Thus, our particular solution is: $f_p(t)=\frac{E_0$$\frac{1}{\omega C}-L\omega$$}{$$\frac{1}{\omega C}-L\omega$$^2+R^2}\cos$$\omega t$$+\frac{E_0R}{$$\frac{1}{\omega C}-L\omega$$^2+R^2}\sin$$\omega t$$$ $f_p(t)=\frac{E_0}{$$\frac{1}{\omega C}-L\omega$$^2+R^2}$$\(\frac{1}{\omega C}-L\omega$$\cos$$\omega t$$+R\sin$$\omega t$$\)$ Using the linear combination identity we may write this as: $f_p(t)=\frac{E_0}{$$\frac{1}{\omega C}-L\omega$$^2+R^2}$$\sqrt{\(\frac{1}{\omega C}-L\omega$$^2+R^2}\sin$$\omega t-\phi$$\)$ where $\tan$$\phi$$=\frac{\omega L-\frac{1}{\omega C}}{R}$ Defining $Z=\sqrt{R^2+$$\omega L-\frac{1}{\omega C}$$^2}$ we may now write: $f_p(t)=\frac{E_0}{Z}\sin$$\omega t-\phi$$$ Now, the general solution f(t) is found by the superposition: $f(t)=f_h(t)+f_p(t)=e^{-\frac{R}{2L}t}$$A\cos(vt)+B\sin(vt)$$+\frac{E_0}{Z }\sin$$\omega t-\phi$$$
 July 7th, 2011, 05:30 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,208 Thanks: 516 Math Focus: Calculus/ODEs Re: Differential equation problem Here are a few words on some of the methods I used. First, consider the following homogeneous equation: $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0$ Because of the nature of the equation, it is reasonable to assume a solution will be of the form: $y(x)=e^{rx}$ where $r\in\mathbb R$ Differentiating with respect to x, we find: $\frac{dy}{dx}=re^{rx}$ $\frac{d^2y}{dx^2}=r^2e^{rx}$ Substituting into the ODE gives: $ar^2e^{rx}+bre^{rx}+ce^{rx}=0$ $e^{rx}$$ar^2+br+c$$=0$ Since $e^{rx}\ne0$ we are left with the auxiliary or characteristic equation: $ar^2+br+c=0$ Thus, by the quadratic formula, we find the roots of the auxiliary equation to be: $r\frac{-b\pm\sqrt{b^2-4ac}}{2a}=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$ If we define: $r_1=-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$ $r_2=-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}$ Then we have two solutions: $y_1(x)=c_1e^{r_1x}$ $y_2(x)=c_2e^{r_2x}$ where $c_i$ are arbitrary constants. Thus, we know: $a\frac{d^2y_1}{dx^2}+b\frac{dy_1}{dx}+cy_1=0$ $a\frac{d^2y_2}{dx^2}+b\frac{dy_2}{dx}+cy_2=0$ Adding the two equations, we get: $a$$\frac{d^2y_1}{dx^2}+\frac{d^2y_2}{dx^2}$$+b$$\f rac{dy_1}{dx}+\frac{dy_2}{dx}$$+c$$y_1+y_2$$=0$ Now, if we define: $y=y_1+y_2$ then $\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}$ $\frac{d^2y}{dx^2}=\frac{d^2y_1}{dx^2}+\frac{d^2y_2 }{dx^2}$ and we may now write: $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0$ where $y(x)=y_1(x)+y_2(x)=c_1e^{r_1x}+c_2e^{r_2x}$ Now, if the roots of the auxiliary equation are complex conjugates, then we may express them as: $r_1=\alpha+\beta i$ $r_2=\alpha-\beta i$ where: $\alpha=-\frac{b}{2a}$ $\beta=\frac{\sqrt{4ac-b^2}}{2a}$ Thus, we may express the solution as: $y(x)=c_1e^{$$\alpha+\beta i$$x}+c_2e^{$$\alpha-\beta i$$x}=e^{\alpha x}$$c_1e^{\beta i\cdot x}+c_2e^{-\beta i\cdot x}$$$ Now, using Euler's formula $e^{ix}=\cos(x)+i\sin(x)$ we may express the solution as: $y(x)=e^{\alpha x}$$c_1\(\cos(\beta x)+i\sin(\beta x)$$+c_2$$\cos(-\beta x)+i\sin(-\beta x)$$\)$ Since $\cos$$-\theta$$=\cos$$\theta$$$ and $\sin$$-\theta$$=-\sin$$\theta$$$ we may write: $y(x)=e^{\alpha x}$$c_1\(\cos(\beta x)+i\sin(\beta x)$$+c_2$$\cos(\beta x)-i\sin(\beta x)$$\)$ $y(x)=e^{\alpha x}$$\(c_1+c_2$$\cos(\beta x)+i$$c_1-c_2$$\sin(\beta x)\)$ Since $c_i$ are arbitrary constants, we may write: $y(x)=e^{\alpha x}$$c_1\cos(\beta x)+ic_2\sin(\beta x)$$$ Now, as we saw above, we know: $y_1(x)=e^{\alpha x}c_1\cos(\beta x)$ $y_2(x)=e^{\alpha x}ic_2\sin(\beta x)$ Are both linearly independent solutions to the ODE, thus if we define: $y_3(x)=\frac{y_2(x)}{i}$ then we have: $a\frac{d^2y_3}{dx^2}+b\frac{dy_3}{dx}+cy_3=0$ $\frac{1}{i}$$a\frac{d^2y_2}{dx^2}+b\frac{dy_2}{dx} +cy_2$$=0$ Since $y_2(x)$ is a solution, then so is $y_3(x)$ and we may express the general solution as: $y(x)=e^{\alpha x}$$c_1\cos(\beta x)+c_2\sin(\beta x)$$$ Now, consider the inhomogeneous equation: $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=g(x)$ and the corresponding homogeneous equation: $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0$ Suppose $y_p(x)$ is a particular solution to the inhomogeneous equation and $y_h(x)$ is the solution to the homogeneous equation. Note that the two solutions must be linearly independent, we then have: $a\frac{d^2y_p}{dx^2}+b\frac{dy_p}{dx}+cy_p=g(x)$ $a\frac{d^2y_h}{dx^2}+b\frac{dy_h}{dx}+cy_h=0$ Adding the two equations, we get: $a$$\frac{d^2y_h}{dx^2}+\frac{d^2y_p}{dx^2}$$+b$$\f rac{dy_h}{dx}+\frac{dy_p}{dx}$$+c$$y_h+y_p$$=g(x)$ With $y(x)=y_h(x)+y_p(x)$ we now have: $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=g(x)$ The particular solution will depend on the form taken by g(x) naturally, and your book should have a table listing the form it will take.
 July 7th, 2011, 05:35 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,885 Thanks: 1088 Math Focus: Elementary mathematics and beyond Re: Differential equation problem Thanks!
 July 7th, 2011, 06:26 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,983 Thanks: 1853 What is represented by the little yellow blob on the left-hand side of the diagram?
 July 7th, 2011, 06:35 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,885 Thanks: 1088 Math Focus: Elementary mathematics and beyond Re: Differential equation problem I don't see it. I don't think it means anything, whatever it is, the diagram is just a resistor, inductor and capacitor with some other component E, connected by lines. There is some discoloration on the lines.
 July 7th, 2011, 07:47 PM #7 Global Moderator   Joined: Dec 2006 Posts: 19,983 Thanks: 1853 I assume the gaps in the wiring lines are similarly unintended. The "other component" you refer to is presumably a power supply.
 July 7th, 2011, 09:03 PM #8 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Nice solution, MarkFL! More info on electromagnetism is here.
 July 7th, 2011, 09:10 PM #9 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,208 Thanks: 516 Math Focus: Calculus/ODEs Re: Differential equation problem Thanks! It's interesting that an under-damped spring-mass system subject to an external sinusoidal force is governed by exactly the same type of ODE. Note that as $t\to\infty$, we have $f_h(t)\rightarrow0$ thus this is called the transient solution, and the other term is called the steady-state solution.
July 8th, 2011, 07:15 AM   #10
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Re:

Quote:
 Originally Posted by skipjack I assume the gaps in the wiring lines are similarly unintended.
Yes.

Quote:
 Originally Posted by skipjack The "other component" you refer to is presumably a power supply.
I don't know. It oscillates.

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