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January 11th, 2018, 10:52 AM   #1
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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?

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Last edited by skipjack; January 11th, 2018 at 09:39 PM.
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January 11th, 2018, 11:16 AM   #2
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Take c=0.
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January 11th, 2018, 11:27 AM   #3
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Originally Posted by Micrm@ss View Post
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?
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January 11th, 2018, 03:50 PM   #4
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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 $y \ne 0$).
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January 11th, 2018, 09:47 PM   #5
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Specifically, y' = y implies e^(-x)y' - e^(-x)y = 0.
Integrating that equation gives e^(-x)y = c, so y = ce^x.
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January 12th, 2018, 11:07 AM   #6
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Quote:
Originally Posted by v8archie View Post
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?
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January 12th, 2018, 11:51 AM   #7
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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.
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Last edited by v8archie; January 12th, 2018 at 11:56 AM.
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January 12th, 2018, 12:38 PM   #8
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That's right, I forgot the absolute value.

Muchas gracias por la ayuda v8archie. Saludos.
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January 12th, 2018, 01:51 PM   #9
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De nada.
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