My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum


Reply
 
LinkBack Thread Tools Display Modes
June 13th, 2015, 06:45 AM   #1
Senior Member
 
Joined: Nov 2010
From: Indonesia

Posts: 1,310
Thanks: 114

[ASK] Solution of Differential Equations

Can anybody help me how to find the solutions of these questions except number 3? The solutions are in the form of x = f(t) and y = f(t).

In the question number 1, I got (D - 1)(D - 4) = 0 whereas $\displaystyle D=\frac{d}{dt}$. Someone please finish this question so that I can use the steps for solving question number 2. Also, how to solve the question number 4 and 5?
Monox D. I-Fly is online now  
 
June 13th, 2015, 08:55 AM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 17,919
Thanks: 1383

(1) Let $u = y + x$, then $du/dt = 4u$, so $u = \text{C}e^{4t}\!$.
$\ \ \ \ \,$ Let $v = 2y - x$, then $dv/dt = v$, etc.

(2) Let $u = x + y$, then $du/dt = 5u$, so $u = \text{C}e^{5t}\!$.
$\ \ \ \ \ \ $Let $v = 3x - 4y$, then $dv/dt = -2v$, etc.
skipjack is offline  
June 13th, 2015, 09:02 AM   #3
Senior Member
 
Joined: Nov 2010
From: Indonesia

Posts: 1,310
Thanks: 114

But the book asks us to use the D = d/dt operator.
Monox D. I-Fly is online now  
June 13th, 2015, 12:08 PM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 17,919
Thanks: 1383

It amounts to the same thing.

For the first question, (D - 3)x - 2y = 0 and (D - 2)y - x = 0.
Adding those gives (D - 4)x + (D - 4)y = 0, i.e., (D - 4)(x + y) = 0,
which is equivalent to du/dt = 4u in my previous post.
skipjack is offline  
June 13th, 2015, 06:32 PM   #5
Senior Member
 
Joined: Nov 2010
From: Indonesia

Posts: 1,310
Thanks: 114

The book also asks to use elimination method, though, so I can't simply add those equations. Oh well, after that part I got ((D - 1)(D - 4) = 0) what are the next steps after we know that D = 1 or D = 4?
Monox D. I-Fly is online now  
June 13th, 2015, 07:39 PM   #6
Senior Member
 
Joined: Nov 2010
From: Indonesia

Posts: 1,310
Thanks: 114

OK, now I understand.
If $\displaystyle \frac{dv}{dt}=v$, is $\displaystyle v=Ce^t$?
Monox D. I-Fly is online now  
June 13th, 2015, 08:00 PM   #7
Global Moderator
 
Joined: Dec 2006

Posts: 17,919
Thanks: 1383

I didn't "simply add". In general, I would need to factorize a quadratic equation to determine what linear combinations to use.

Using your method, you got D² - 5D + 4)x = 0 or D² - 5D + 4)y = 0. You then factorized (the quadratic in D) to get (D - 1)(D - 4)x = 0.

You cannot conclude that D = 1 or D = 4. What you have is simply a second order linear ordinary differential equation (with constant coefficients). The factorization has already been done, so you can write down its solution as a linear combination of e^t and e^(4t). Having found x (say), you can substitute for x in the first equation you were given and hence find y.
skipjack is offline  
June 13th, 2015, 08:16 PM   #8
Senior Member
 
Joined: Nov 2010
From: Indonesia

Posts: 1,310
Thanks: 114

So, x = e^t? Substituting it in the first equation, I will get:
e^t = 3e^t + 2y
-2e^t = 2y
y = -e^t

Like that?

And where did you get v = 2y - x in the first question?

Is the solution $\displaystyle x = e^{4t}$ and $\displaystyle x = e^t$ or $\displaystyle x = c_1e^{4t} +c_2e^t$?

Last edited by skipjack; June 14th, 2015 at 02:09 AM.
Monox D. I-Fly is online now  
June 14th, 2015, 02:14 AM   #9
Global Moderator
 
Joined: Dec 2006

Posts: 17,919
Thanks: 1383

$\displaystyle x = c_1e^{4t} + c_2e^t$

If you substitute that in the first equation, what do you get?
skipjack is offline  
June 14th, 2015, 02:28 AM   #10
Senior Member
 
Joined: Nov 2010
From: Indonesia

Posts: 1,310
Thanks: 114

Let's see:
$\displaystyle x=c_1e^{4t}+c_2e^t$
$\displaystyle \frac{dx}{dt}=4c_1e^{4t}+c_2e^t$

$\displaystyle \frac{dx}{dt}=3x+2y$
$\displaystyle 4c_1e^{4t}+c_2e^t=3(c_1e^{4t}+c_2e^t)+2y$
$\displaystyle 4c_1e^{4t}+c_2e^t=3c_1e^{4t}+3c_2e^t)+2y$
$\displaystyle 2y=c_1e^{4t}-2c_2e^t$
$\displaystyle y=\frac{1}{2}c_1e^{4t}-c_2e^t$

Am I right?
Monox D. I-Fly is online now  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
differential, equations, solution



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Differential Equations: General & Particular Solution aliaja Differential Equations 1 September 17th, 2013 03:31 PM
Differential Equations Verifying Implicit Solution aztakurva Differential Equations 2 August 22nd, 2013 06:15 PM
differential equations solution - help! science Differential Equations 2 July 17th, 2013 08:08 AM
The system of equations (64 equations -12 unknown) solution ZeusTheMunja Linear Algebra 20 January 25th, 2013 05:52 AM
how to find a solution the differential equations mulyak.oleksandr Differential Equations 1 September 5th, 2012 06:17 AM





Copyright © 2017 My Math Forum. All rights reserved.