My Math Forum [ASK] Solution of Differential Equations

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 June 13th, 2015, 06:45 AM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,962 Thanks: 132 Math Focus: Trigonometry [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, 08:55 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,066 Thanks: 1621 (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, 09:02 AM #3 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,962 Thanks: 132 Math Focus: Trigonometry But the book asks us to use the D = d/dt operator.
 June 13th, 2015, 12:08 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,066 Thanks: 1621 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, 06:32 PM #5 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,962 Thanks: 132 Math Focus: Trigonometry 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, 07:39 PM #6 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,962 Thanks: 132 Math Focus: Trigonometry OK, now I understand. If $\displaystyle \frac{dv}{dt}=v$, is $\displaystyle v=Ce^t$?
 June 13th, 2015, 08:00 PM #7 Global Moderator   Joined: Dec 2006 Posts: 19,066 Thanks: 1621 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, 08:16 PM #8 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,962 Thanks: 132 Math Focus: Trigonometry 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.
 June 14th, 2015, 02:14 AM #9 Global Moderator   Joined: Dec 2006 Posts: 19,066 Thanks: 1621 $\displaystyle x = c_1e^{4t} + c_2e^t$ If you substitute that in the first equation, what do you get?
 June 14th, 2015, 02:28 AM #10 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,962 Thanks: 132 Math Focus: Trigonometry 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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