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### Messages - Julian

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##### Chapter 2 / W2L3 What's u in the general solution to a homogeneous equation?
« on: September 28, 2020, 12:57:40 PM »
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.)

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##### Chapter 2 / W3L3 Exact solutions to inexact equations
« on: September 28, 2020, 12:45:14 PM »
In week 3 lecture 3, we get the example $(-y\sin(x)+y^3\cos(x))dx+(3\cos(x)+5y^2\sin(x))dy=0$. We determine that this equation is not exact, but that we can make it exact by multiplying the equation by $y^2$. We then find the general solution to the new equation is $y^3\cos(x)+y^5\sin(x)=C$.

My question is why is this good enough? We didn't answer the original question. We answered a modified version which we chose specifically because it seems easier to us. Shouldn't we still find a solution to the original equation $(-y\sin(x)+y^3\cos(x))dx+(3\cos(x)+5y^2\sin(x))dy=0$? Is there some way we can "divide out" $y^2$ from $y^3\cos(x)+y^5\sin(x)=C$ to get it?

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