My Math Forum Diff Eq - Solve the initial value problem

 Calculus Calculus Math Forum

 September 12th, 2013, 06:54 PM #1 Member   Joined: Jun 2012 From: San Antonio, TX Posts: 84 Thanks: 3 Math Focus: Differential Equations, Mathematical Modeling, and Dynamical Systems Diff Eq - Solve the initial value problem $3y' -6y=e^{-\frac{\pi}{2}t}$, $y(0)=a$ Let $a_0$ be the value of $a$ for which the transition from one type of behavior to another occurs. Find the critical value $a_0$ exactly. I've transformed the equation into $y'-2y=\frac{1}{3}e^{-\frac{\pi t}{2}}$ so, $p(t)=-2$ and $\mu(t)=e^{-2t}$ I multiply by integrating factor: $e^{-2t}y' -2e^{-2t}y=\frac{1}{3}e^{-\frac{\pi t}{2}}*e^{-2t}$ Rewrite as: $d\left[e^{-2t}y\right]=\left[\frac{1}{3}e^{-\frac{\pi t}{2}}*e^{-2t}\right]dt$ Then: $e^{-2t}y=\int\left[\frac{1}{3}e^{-\frac{\pi t}{2}}*e^{-2t}\right]dt=e^{-2t}y=\frac{1}{3}\int\left[e^{\frac{t(-\pi -4)}{2}}\right]dt$ $e^{-2t}y=\frac{1}{3}\left[\frac{2}{-\pi -4}e^{\frac{t(-\pi -4)}{2}}\right]+C$ $y=\frac{1}{3}e^{2t}\left[\frac{2}{-\pi -4}e^{\frac{t(-\pi -4)}{2}}+C\right]$ $a=\frac{2}{-\pi -4}+C$ and $C=a-\frac{2}{-\pi -4}$ I feel like I've made this too complicated to find $a_0$ Can someone give me some tips
 September 12th, 2013, 07:44 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Diff Eq - Solve the initial value problem Using essentially the same method, I get: $y(t)=-\frac{2}{3(4+\pi)}e^{-\frac{\pi}{2}t}+$$a+\frac{2}{3(4+\pi)}$$e^{2t}$
September 12th, 2013, 09:39 PM   #3
Member

Joined: Jun 2012
From: San Antonio, TX

Posts: 84
Thanks: 3

Math Focus: Differential Equations, Mathematical Modeling, and Dynamical Systems
Re: Diff Eq - Solve the initial value problem

Quote:
 Originally Posted by MarkFL Using essentially the same method, I get: $y(t)=-\frac{2}{3(4+\pi)}e^{-\frac{\pi}{2}t}+$$a+\frac{2}{3(4+\pi)}$$e^{2t}$
How were you able to make it look so... neat?

I got the question correct by guessing, but I'm still uncertain on how I'm supposed to find the critical values. I'm assuming you use the first derivative test, but I can't seem to find a zero or an undefined point.

 September 12th, 2013, 10:25 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Diff Eq - Solve the initial value problem This is my working: Like you I wrote the ODE in the form: $y'-2y=\frac{1}{3}e^{-\frac{\pi}{2}t}$ where $y(0)=a$ We find the integrating factor is: $\mu(t)=e^{-2\int\,dt}=e^{-2t}$ and so the ODE becomes: $e^{-2t}y'-2e^{-2t}y=\frac{1}{3}e^{-\frac{\pi}{2}t}e^{-2t}=\frac{1}{3}e^{-\frac{4+\pi}{2}t}$ Writing the left side as the differentiation of a product, we have: $\frac{d}{dt}$$e^{-2t}y$$=\frac{1}{3}e^{-\frac{4+\pi}{2}t}$ Integrating with respect to $t$, we find: $e^{-2t}y=-\frac{2}{3(4+\pi)}e^{-\frac{4+\pi}{2}t}+C$ Multiply through by $e^{2t}$: $y(t)=-\frac{2}{3(4+\pi)}e^{-\frac{\pi}{2}t}+Ce^{2t}$ Now we may using the initial value to determine the value of the parameter $C$: $y(0)=-\frac{2}{3(4+\pi)}+C=a\,\therefore\,C=a+\frac{2}{3 (4+\pi)}$ And so the solution satisfying the IVP is: $y(t)=-\frac{2}{3(4+\pi)}e^{-\frac{\pi}{2}t}+$$a+\frac{2}{3(4+\pi)}$$e^{2t}$
September 13th, 2013, 09:24 AM   #5
Member

Joined: Jun 2012
From: San Antonio, TX

Posts: 84
Thanks: 3

Math Focus: Differential Equations, Mathematical Modeling, and Dynamical Systems
Re: Diff Eq - Solve the initial value problem

Quote:
 Originally Posted by MarkFL And so the solution satisfying the IVP is: $y(t)=-\frac{2}{3(4+\pi)}e^{-\frac{\pi}{2}t}+$$a+\frac{2}{3(4+\pi)}$$e^{2t}$
I'm still uncertain on how to find $a_0$

The hint that I'm given is this:

Find the general solution y(t) and solve for y'(0) = 0 . The initial value a0 obtained from this equation will be the critical value separating the solutions which increase without bound with t ? 0 from those which decrease without bound with t ? 0.

The answer I typed in is $a_0=-\frac{2}{3(4+\pi )$ but I don't know how to come to that answer

 September 13th, 2013, 09:46 AM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Diff Eq - Solve the initial value problem This makes little sense to me. As $t\to\0$ we should expect $y(t)\to a$. It is the behavior as $t\to\infty$ that is affected by the choice of $a$.
September 13th, 2013, 11:02 AM   #7
Member

Joined: Jun 2012
From: San Antonio, TX

Posts: 84
Thanks: 3

Math Focus: Differential Equations, Mathematical Modeling, and Dynamical Systems
Re: Diff Eq - Solve the initial value problem

Quote:
 Originally Posted by MarkFL This makes little sense to me. As $t\to\0$ we should expect $y(t)\to a$. It is the behavior as $t\to\infty$ that is affected by the choice of $a$.
Alright. You've been a big help. Thanks a lot

 Tags diff, initial, problem, solve

,

,

,

,

,

,

,

,

,

,

,

### differential critical value

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post r-soy Calculus 2 November 2nd, 2012 04:37 AM reggea07 Applied Math 1 May 19th, 2011 02:56 PM r-soy Calculus 18 December 21st, 2010 01:46 PM zgonda Calculus 1 November 27th, 2010 04:23 PM kien Applied Math 0 April 12th, 2009 09:44 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top