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June 13th, 2015, 07:45 AM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,433 Thanks: 118  [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? 
June 13th, 2015, 09:55 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,154 Thanks: 1418 
(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. 
June 13th, 2015, 10:02 AM  #3 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,433 Thanks: 118 
But the book asks us to use the D = d/dt operator. 
June 13th, 2015, 01:08 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,154 Thanks: 1418 
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. 
June 13th, 2015, 07:32 PM  #5 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,433 Thanks: 118 
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?

June 13th, 2015, 08:39 PM  #6 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,433 Thanks: 118 
OK, now I understand. If $\displaystyle \frac{dv}{dt}=v$, is $\displaystyle v=Ce^t$? 
June 13th, 2015, 09:00 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,154 Thanks: 1418 
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. 
June 13th, 2015, 09:16 PM  #8 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,433 Thanks: 118 
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 03:09 AM. 
June 14th, 2015, 03:14 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 18,154 Thanks: 1418 
$\displaystyle x = c_1e^{4t} + c_2e^t$ If you substitute that in the first equation, what do you get? 
June 14th, 2015, 03:28 AM  #10 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,433 Thanks: 118 
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? 

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