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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 January 11th, 2018, 10:52 AM #1 Newbie   Joined: Apr 2015 From: Lima Posts: 21 Thanks: 2 Singular solution of y'=y By solving the DE y'=y I got that y=ce^x and y must not be zero. Then I think that y=0 meet the original equation since y'=0=y. Question: can I consider y= 0 a singular solution that can not be obtained from the general solution y=ce^x? Regards. Last edited by skipjack; January 11th, 2018 at 09:39 PM. January 11th, 2018, 11:16 AM #2 Senior Member   Joined: Oct 2009 Posts: 863 Thanks: 328 Take c=0. January 11th, 2018, 11:27 AM   #3
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Quote:
 Originally Posted by Micrm@ss Take c=0.
In the previous step for solving the equaiton I have to make y'/y = 1

In that moment y cannot be zero. So I think the solution obtained in further steps cannot include y=0.

Am I right? January 11th, 2018, 03:50 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra In dividing by $y$ you have indeed excluded $y=0$ from your family of solutions, but that doesn't stop it from being a solution to the equation. What you have actually done is assumed that $y \ne 0$ before dividing by $y$. And, under the assumption that $y \ne 0$, it is obvious that $y \ne 0$. In order to get a complete solution to the equation, you must notice that your analysis makes the assumption that $y \ne 0$ and therefore go back and see whether there are any more solutions if that assumption doesn't hold. This you have also done by noticing that $y=0$ is indeed a solution. Thus the complete solution of the equation is the union of the two sets of solutions $y = ce^x$ (where $c \ne 0$) and $y=0$. This can be simplified to $$y = c e^x \quad c \in \mathbb R$$ (There are alternative methods of solving the equation that do not require one to assume that $y \ne 0$). Thanks from Snair January 11th, 2018, 09:47 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2216 Specifically, y' = y implies e^(-x)y' - e^(-x)y = 0. Integrating that equation gives e^(-x)y = c, so y = ce^x. January 12th, 2018, 11:07 AM   #6
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Quote:
 Originally Posted by v8archie In dividing by $y$ you have indeed excluded $y=0$ from your family of solutions, but that doesn't stop it from being a solution to the equation. What you have actually done is assumed that $y \ne 0$ before dividing by $y$. And, under the assumption that $y \ne 0$, it is obvious that $y \ne 0$. In order to get a complete solution to the equation, you must notice that your analysis makes the assumption that $y \ne 0$ and therefore go back and see whether there are any more solutions if that assumption doesn't hold. This you have also done by noticing that $y=0$ is indeed a solution. Thus the complete solution of the equation is the union of the two sets of solutions $y = ce^x$ (where $c \ne 0$) and $y=0$. This can be simplified to $$y = c e^x \quad c \in \mathbb R$$ (There are alternative methods of solving the equation that do not require one to assume that $y \ne 0$).
But But but... if c belongs to R that implies y could be negative.

While integrating i get ln y = ... this implies that y must be always greater than zero.

Am I right? January 12th, 2018, 11:51 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra No. You forgot the absolute value signs in the logarithm. You should get \begin{align*}\ln{|y|} &= x + c_1 & \text{(for any value of $c_1$)} \\ |y| &= e^{x+c_1} = e^{c_1}e^x = c_2e^x & \text{(where $c_2 = e^{c_1} \gt 0$)} \\ y &= \pm c_2 e^x & (c_2 \gt 0) \end{align*} We can write this more simply as $$y = c e^x \quad (c \ne 0)$$ And then we add the solution for $c=0$ as described above. Thanks from JoseTorero Last edited by v8archie; January 12th, 2018 at 11:56 AM. January 12th, 2018, 12:38 PM #8 Newbie   Joined: Apr 2015 From: Lima Posts: 21 Thanks: 2 That's right, I forgot the absolute value. Muchas gracias por la ayuda v8archie. Saludos. Thanks from v8archie January 12th, 2018, 01:51 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra De nada. Tags singular, solution Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post eulid Differential Equations 2 November 27th, 2017 10:17 AM levanduong123200 Differential Equations 0 August 31st, 2014 11:31 AM triplekite Calculus 0 October 29th, 2012 02:12 PM mr_dude32 Calculus 1 December 10th, 2008 07:16 PM mr_dude32 Linear Algebra 1 December 10th, 2008 12:15 PM

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