MAT244--2020F > Chapter 2

W2L3 What's u in the general solution to a homogeneous equation?

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Julian:
In tutorial with Neall last week, we were asked to find the general solution to the equation $(xy +y^2+x^2)dy-x^2dx=0$. Based on techniques from class (letting $y=ux$), I determined that the parametric general solution to this (homogeneous) equation is $x=Ce^{\arctan(u)}$ and $y=Cxe^{\arctan(u)}$ or equivalently $x=Ce^{\arctan(\tfrac{y}{x})}$ and $y=Cxe^{\arctan(\tfrac{y}{x})}$.

My question is why is this satisfactory? Shouldn't we find a function $y$ purely in terms of $x$ instead? Indeed, in tutorial, we were suggested that $y=\tan(\ln(Cx))x$. It's really not obvious to me that these are even equivalent. (If anyone can tell me why these are equivalent, I would also appreciate that.)

Victor Ivrii:
It was explained in the lectures that solutions in the parametric form are also admissible. You could ask it on the tutorial.

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