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April 27th, 2015, 01:16 PM  #1 
Newbie Joined: Jan 2014 Posts: 19 Thanks: 0  General solution
Should we leave the general solution of the differential equation $t\frac{dy}{dx}+2y=4t^2$ as $t^2y=t^4+c$ instead of $y=t^2+c/(t^2)$ ($c$ is an arbitrary constant)? Does the solution $y=t^2+c/(t^2)$ include the particular solution $y=t^2$? Should we make the assumption $t\neq0$ before we simplify the solution $t^2y=t^4+c$ to $y=t^2+c/(t^2)$?
Last edited by skipjack; April 27th, 2015 at 04:53 PM. 
April 27th, 2015, 04:51 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
If one is asked to solve a differential equation without being told what the domain of the solutions has to be, one has to consider whether the domain should be considered as implied or whether it will depend on the value of the constant of integration. If the differential equation you gave was presented in a context where the solver may have little or no knowledge of complex numbers, it might be reasonable to suppose that $t$ is real. If one further assumes that $t$ can have any real value, the equation has only one solution, $y = t^2$. This is also the only solution if $t$ is allowed to have any complex value. This is a singular solution, as it's not a special case of the general solution. The differential equation you gave has, in general, a single fixed singularity at $t = 0$. By that, I mean that any solution you find that involves "c" (as a constant of integration) doesn't define y at one value of $t$ (zero in this case) that doesn't depend on the value of c. If the domain had been specified as $t$ > 0, the general solution would have been $y = t^2 + c/t^2$ and as $y = t^2$ is included in that solution, it wouldn't be a singular solution. If $t$ is assumed to be real, could one then say that $y = t^2 + c/t^2$ with domain $t \ne 0$ is the general solution? Strictly speaking, that shouldn't be done, as it excludes the possibility that c has one value for $t$ < 0 and another value for $t$ > 0. Leaving the solution as $t^2y = t^4 + c$ doesn't avoid such difficulties. 

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