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November 7th, 2017, 05:49 AM  #1 
Newbie Joined: Apr 2016 From: Wonderland Posts: 16 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  (x2) 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, 06:32 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,306 Thanks: 2443 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 $x$ ($0  (x2)0 = 0$), so $y=0$ is indeed a solution. 
November 7th, 2017, 07:46 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,306 Thanks: 2443 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, 08:36 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,956 Thanks: 1603 
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  (x2) dy/dx = 0, where x and y are real. 
November 7th, 2017, 10:23 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,306 Thanks: 2443 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  (x2)y' &= 0 \\ y' + \frac{2}{2x} y &= 0 &&& \int \frac{2}{2x} \,\mathrm dx & = 2\ln{(2x)} = \ln{\frac1{(x2)^2}} \\ &&&& p(x) &= e^{\ln{\frac1{(x2)^2}}} = \frac1{(x2)^2} \\ \frac1{(x2)^2}y'  \frac2{(x2)^3}y &= 0 \\ \frac1{(x2)^2} y &= c \\ y &= c(x2)^2 \end{align*} which gives $y=0$ when $c=0$. Alternatively, the both equations can be solved by separating variables. \begin{align*} y  (x2)y' &= 0 \\ \int \frac1y \frac{\mathrm dy}{\mathrm dx}\, \mathrm dx &= \int \frac1{x2} \,\mathrm dx \\ \ln y &= \ln{Ax2} & (A \gt 0) \\ y &= \pm A(x2) & (A \gt 0) \\ y &= c(x2) & (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 $(y2)  (x2)y' = 0$ (solving by separating variables) we divide by $(y2)$ and so we'd have to verify $y2=0 \implies y=2$ manually. Last edited by v8archie; November 7th, 2017 at 10:46 AM. 
November 7th, 2017, 10:34 AM  #6 
Senior Member Joined: Jun 2015 From: England Posts: 822 Thanks: 243 
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, 10:46 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,306 Thanks: 2443 Math Focus: Mainly analysis and algebra 
Quote the post to see the $\LaTeX$.


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