My Math Forum God help me (Differential Equations)

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 January 30th, 2011, 02:29 PM #1 Member   Joined: Oct 2009 Posts: 98 Thanks: 0 God help me (Differential Equations) I have this problem that is on our homework that is due Monday. We have had snow days and therefore we have not had our homework adjusted to the rate at which we are learning due to the snow closures. So as a result this problem is in out homework and I have NO clue how to do this. We have NOT even started second order differential equations and we are expected to have this done by tomorrow. Please help me. And thank you for your time in advance. Because this will take time. This is what you get when your math teacher has a Ph. D. in mathematics from Princeton. Lol Uploaded with ImageShack.us
 January 30th, 2011, 04:08 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: God help me (Differential Equations) We are given: $\frac{d^2y}{dt^2}-4\frac{dy}{dt}+4y=$$1+t+t^2+\cdots+t^{29}$$e^{2t}$ We can show, using the method of variation of parameters, that the expression given in the hint is a particular solution, to which we can then add the general solution to the corresponding homogeneous equation to get a general solution. First we find a fundamental solution set $\{y_1(t),y_2(t)\}$ for the corresponding homogeneous equation and take: $y_p(t)=v_1(t)y_1(t)+v_2(t)y_2(t)$ The auxiliary equation is $r^2-4r+4=(r-2)^2=0$ and since we have a repeated root, we now have $y_p(t)=v_1(t)te^{2t}+v_2(t)e^{2t}$ Next, we determine $v_1(t)$ and $v_2(t)$ by solving the system: $te^{2t}v_1'+e^{2t}v_2'=0$ $(2t+1)e^{2t}v_1'+2e^{2t}v_2'=$$1+t+t^2+\cd ots+t^{29}$$e^{2t}$ First, we can divide both equations by $e^{2t}$ since it can never be zero, to get: $tv_1'+v_2'=0$ $(2t+1)v_1'+2v_2'=$$1+t+t^2+\cdots+t^{29}$$$ Solving the system, we find: $v_1'=$$1+t+t^2+\cdots+t^{29}$$$ $v_2'=-$$t+t^2+t^3+\cdots+t^{30}$$$ Now, integrating to find $v_1(t)$ and $v_2(t)$ yields: $v_1(t)=$$t+\frac{t^2}{2}+\frac{t^3}{3}+\cdots+\fra c{t^{30}}{30}$$+C_1$ $v_2(t)=-$$\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^4}{4}+\cdots +\frac{t^{31}}{31}$$+C_2$ Since we need only one particular solution, we take $C_1=C_2=0$ for simplicity. Now we substitute for $v_1(t)$ and $v_2(t)$ into the expression for $y_p(t)$ to obtain a particular solution. $y_p(t)=$$t+\frac{t^2}{2}+\frac{t^3}{3}+\cdots+\fra c{t^{30}}{30}$$te^{2t}-$$\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^4}{4}+\cdots +\frac{t^{31}}{31}$$e^{2t}$ Now, factoring $t^2e^{2t}$ from both terms gives: $y_p(t)=t^2$$$1-\frac{1}{2}$$+$$\frac{1}{2}-\frac{1}{3}$$t+$$\frac{1}{3}-\frac{1}{4}$$t^2+\cdots+$$\frac{1}{30}-\frac{1}{31}$$t^{29}$e^{2t}$ Or, written more efficiently using sigma notation: $y_p(t)=t^2$\sum_{n=1}^{30}$$\frac{1}{n(n+1)}t^{n-1}$$$e^{2t}$ Recall that a general solution to a non-homogeneous equation is given by the sum of a general solution to the homogeneous equation and a particular solution. Consequently, a general solution to the given equation is: $y(t)=y_h(t)+y_p(t)=c_1te^{2t}+c_2e^{2t}+t^2$\sum_{n=1}^{30}$$\frac{1}{n(n+1)}t^{n-1}$$$e^{2t}$
 January 30th, 2011, 04:21 PM #3 Member   Joined: Oct 2009 Posts: 98 Thanks: 0 Re: God help me (Differential Equations) MarkFL, my hat tips to you. I am so thankful for your help and time!! I feel bad posting questions that I haven't taken a try at, but for this one, I was really unable to even start it. We have not had lectures on this material due to snow days and the homework packets have not been adjusted to the lecture schedule resulting in material being on the homework that we have not learned. I am so grateful for your help. So grateful. *hat tips*
 January 30th, 2011, 04:27 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: God help me (Differential Equations) Yeah, I can't imagine being expected to solve such an equation without having had some lectures on the theory of linear 2nd order ODEs. It is possible there is a much simpler way to solve this equation, so check back as someone may know a more efficient way to go about it.
 February 2nd, 2011, 06:24 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,101 Thanks: 1905 e^(-2t)y'' - 4e^(-2t)y' + 4e^(-2t)y = 1 + t + t² + ... + t^29 e^(-2t)y' - 2e^(-2t)y = t + t²/2 + t³/3 + ... + t^30/30 + A e^(-2t)y = t²/2 + t³/(2*3) + ... + t^31/(30*31) + At + B y = (t²/2 + t³/(2*3) + ... + t^31/(30*31) + At + B)e^(2t), where A and B are constants.
 February 2nd, 2011, 10:55 AM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: God help me (Differential Equations) Ah, the "someone" I had in mind shows how it's done. First, both sides were multiplied by $e^{-2t}$ to get: $e^{-2t}\frac{d^2y}{dt^2}-4e^{-2t}\frac{dy}{dt}+4e^{-2t}y=1+t+t^2+\cdots+t^{29}$ Then, the recognition that $\frac{d}{dt}$$e^{-2t}f(t)$$=e^{-2t}\frac{df}{dt}-2e^{-2t}f$ was used twice to write the equation as: $$$e^{-2t}\frac{d^2y}{dt^2}-2e^{-2t}\frac{dy}{dt}$$-2$$e^{-2t}\frac{dy}{dt}-2e^{-2t}y$$=1+t+t^2+\cdots+t^{29}$ $\frac{d}{dt}$$e^{-2t}\frac{dy}{dt}$$-2\frac{d}{dt}$$e^{-2t}y$$=1+t+t^2+\cdots+t^{29}$ $\frac{d}{dt}$$\frac{d}{dt}\(e^{-2t}y$$\)=1+t+t^2+\cdots+t^{29}$ Integrating twice: $\int\int\,d^2$$e^{-2t}y$$=\int\int 1+t+t^2+\cdots+t^{29}\,dt\,dt$ $e^{-2t}y=\frac{t^2}{2}+\frac{t^3}{2\cdot3}+\frac{t^4}{ 3\cdot4}+\cdots+\frac{t^{31}}{30\cdot31}+At+B$ $y(t)=$$\frac{t^2}{2}+\frac{t^3}{2\cdot3}+\frac{t^4 }{3\cdot4}+\cdots+\frac{t^{31}}{30\cdot31}+At+B$$e ^{2t}$
February 2nd, 2011, 11:05 AM   #7
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Re: God help me (Differential Equations)

Quote:
 Originally Posted by MarkFL Ah, the "someone" I had in mind shows how it's done. ...
Yeah, I knew there had to be a better (i.e. faster) way!

February 2nd, 2011, 11:09 AM   #8
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Re: God help me (Differential Equations)

Quote:
 Originally Posted by The Chaz ... Yeah, I knew there had to be a better (i.e. faster) way!
That was my suspicion as well, but I couldn't make the leap. My hat tips to [color=#008000]skipjack[/color]!

In retrospect, a simple "check" of the solution would have easily shown me, but hindsight...

 February 2nd, 2011, 11:47 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,101 Thanks: 1905 I used the same method here, and you acknowledged you had read it.
 February 2nd, 2011, 12:08 PM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: God help me (Differential Equations) Yes, I remember now. I guess old habits die hard, but I will make an honest effort to remember this technique in the future, as it certainly is a time saver. Thank you for demonstrating a more efficient approach.

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