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 November 7th, 2017, 06:49 AM #1 Newbie   Joined: Apr 2016 From: Wonderland Posts: 14 Thanks: 0 How to tell when y = 0 is a solution? Hi there, when solving differential equations, how do you know when y = 0 is a solution? For example: Solve the differential equation y - (x-2) dy/dx = 0 I don't really understand the how "y = 0" can be a solution. Any guides? And is this applicable for first order & second order (non) homogeneous equations?
 November 7th, 2017, 07:32 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,137 Thanks: 2381 Math Focus: Mainly analysis and algebra If $y=0$ is a solution, it is constant. So $\frac{\mathrm dy}{\mathrm dx}=0$. Plugging into the given equation, we see that it is satisfied for all values of $x$ ($0 - (x-2)0 = 0$), so $y=0$ is indeed a solution. Thanks from topsquark
 November 7th, 2017, 08:46 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,137 Thanks: 2381 Math Focus: Mainly analysis and algebra This approach works for all equations because all we are doing is checking whether or not a given solution satisfies the given equation.
 November 7th, 2017, 09:36 AM #4 Global Moderator   Joined: Dec 2006 Posts: 18,554 Thanks: 1479 This overlooks why one would be interested in testing y = 0 in the first place. I suggest taking a further look at the slightly different ODE, 2y - (x-2) dy/dx = 0, where x and y are real. Thanks from topsquark
 November 7th, 2017, 11:23 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,137 Thanks: 2381 Math Focus: Mainly analysis and algebra I'm not sure what you are getting at there Skipjack. The two equations can be solved in exactly the same ways. \begin{align*} 2y - (x-2)y' &= 0 \\ y' + \frac{2}{2-x} y &= 0 &&& \int \frac{2}{2-x} \,\mathrm dx & = -2\ln{(2-x)} = \ln{\frac1{(x-2)^2}} \\ &&&& p(x) &= e^{\ln{\frac1{(x-2)^2}}} = \frac1{(x-2)^2} \\ \frac1{(x-2)^2}y' - \frac2{(x-2)^3}y &= 0 \\ \frac1{(x-2)^2} y &= c \\ y &= c(x-2)^2 \end{align*} which gives $y=0$ when $c=0$. Alternatively, the both equations can be solved by separating variables. \begin{align*} y - (x-2)y' &= 0 \\ \int \frac1y \frac{\mathrm dy}{\mathrm dx}\, \mathrm dx &= \int \frac1{x-2} \,\mathrm dx \\ \ln |y| &= \ln{A|x-2|} & (A \gt 0) \\ y &= \pm A(x-2) & (A \gt 0) \\ y &= c(x-2) & (c \ne 0) \end{align*} And in this case we don't get $y=0$ because $c=0$ is not permitted (because the logarithm of zero is not defined). We lost the solution $y=0$ at line 2 where we divided by $y$ because this is only possible when $y \ne 0$. Thus we manually verify whether or not $y=0$ is a solution as above. In the equation $(y-2) - (x-2)y' = 0$ (solving by separating variables) we divide by $(y-2)$ and so we'd have to verify $y-2=0 \implies y=2$ manually. Thanks from topsquark Last edited by v8archie; November 7th, 2017 at 11:46 AM.
 November 7th, 2017, 11:34 AM #6 Senior Member   Joined: Jun 2015 From: England Posts: 732 Thanks: 207 Archie, I would be very grateful to learn how you managed to lay that math out on the page like you did.
 November 7th, 2017, 11:46 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,137 Thanks: 2381 Math Focus: Mainly analysis and algebra Quote the post to see the $\LaTeX$.

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