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June 17th, 2018, 12:47 PM  #1 
Member Joined: Jul 2014 From: Amherst, MA Posts: 34 Thanks: 6 Math Focus: Calculus, Differential Geometry, Physics, Topology  Question: Getting rid of Constants?
I was taking a physics class once, my prof was solving a PDE, and he did some trick that got rid of all the constants in the equation. For example, if he had something like $\displaystyle \alpha y_{tt}  \beta y_{xx} = 0$, he would end up with something like $\displaystyle u_{tt}u_{xx}=0$, then after he solved the thing, he would substitute back, and get the solution to the first equation. Can you do this with any equation? I don't know if the method depended on some other property like linearity, I wasn't really paying attention. He did it at least under some conditions, and I can't figure out how to do it. Any ideas? Could make some work cleaner and easier to keep straight. Russ Sent from my SMG955U using Tapatalk 
June 18th, 2018, 03:52 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
I can't say for sure what your professor did but I would take, not a new dependent variable, but two new independent variables, $\displaystyle x'= \frac{x}{\sqrt{\alpha}}$ and $\displaystyle t'= \frac{t}{\sqrt{\beta}}$. Then, by the chain rule, $\displaystyle \frac{\partial y}{\partial x}= \frac{\partial y}{\partial x'}\frac{dx'}{dx}= \frac{\frac{\partial y}{\partial x'}}{\sqrt{\alpha}}$ and then $\displaystyle \frac{\partial^2 y}{\partial x^2}= \frac{\partial}{\partial x}\left(\frac{\partial y}{\partial x}\right)= \frac{\frac{\partial}{\partial x'}}{\sqrt{\alpha}}= \frac{\frac{\partial^2 y}{\partial x'^2}}{\alpha}$. Similarly for t. 

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