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Chapter 2 / Re: Week 2 Lec 1 (Chapter 2) question
« on: January 17, 2022, 07:48:38 PM »
Now it is correct $x=Ce^{t}$ and then $C=?$
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In particular, the definition of a linear PDE, from the textbook, is: $au_{x}+bu_{y}+cu-f=0$, where $f= f(x,y)$. However, if we simply move the the $cu$ to the right-hand side, we get: $au_{x}+bu_{y}=f-cu$. Now, define $g(x,y,u) = f(x,y)-cu$, then $au_{x}+bu_{y}=g(x,y,u)$, and the right-hand side now depends on lower-order derivatives, so by definition, it's quasi-linear. Could someone help identify the issue with this argument?First, it will be not just quasilinear, but also semilinear. Second, it will also be linear since you can move $c(x,y)u$ to the left