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January 11th, 2018, 10:52 AM  #1 
Newbie Joined: Apr 2015 From: Lima Posts: 10 Thanks: 1  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: 275 Thanks: 92 
Take c=0.

January 11th, 2018, 11:27 AM  #3 
Newbie Joined: Apr 2015 From: Lima Posts: 10 Thanks: 1  
January 11th, 2018, 03:50 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,234 Thanks: 2412 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 $y \ne 0$). 
January 11th, 2018, 09:47 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,704 Thanks: 1530 
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  
Newbie Joined: Apr 2015 From: Lima Posts: 10 Thanks: 1  Quote:
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: 7,234 Thanks: 2412 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. 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: 10 Thanks: 1 
That's right, I forgot the absolute value. Muchas gracias por la ayuda v8archie. Saludos. 
January 12th, 2018, 01:51 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,234 Thanks: 2412 Math Focus: Mainly analysis and algebra 
De nada.


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