anonymous
  • anonymous
can anyone help me how to solve this homogenous equation please: [x csc (y/x) - y] dx + x dy = 0 ..its says here that the answer should be ln l x/c l = cos (y/x)
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amoodarya
  • amoodarya
take \[\frac{y}{x}=v\\y=vx\\y'=v+xv'\\(xcsc(\frac{y}{x})-y)dx+xdy=0\\(xcsc(\frac{y}{x})-y)+x\frac{dy}{dx}=0\\ divx\\(\csc(\frac{y}{x})-\frac{y}{x})+y'=0\\\]
amoodarya
  • amoodarya
if you put v it will be simple to solve can you go on ?
UsukiDoll
  • UsukiDoll
this is substitution method for first order odes

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anonymous
  • anonymous
i'll try :) our prof didn't really explained it to us very well the steps in solving it
UsukiDoll
  • UsukiDoll
usually you have to do substitution and then argh I forgot... a. I didn't have trig involved for my odes b. substitution was my least favorite.
anonymous
  • anonymous
actually i have already taken up integral and differential calculus but this is my first time encountering this topic
UsukiDoll
  • UsukiDoll
I really didn't like this method at all when I came across it x(
UsukiDoll
  • UsukiDoll
my favorites are integrating factor and exact oh yeah!
amoodarya
  • amoodarya
\[(\csc(v)-v)+(v+v'x)=0\\csc(v)+xv'=0\\xv'=-\csc(v)\\x\frac{dv}{dx}=\frac{1}{\sin(v)}\\-\sin(v)dv=\frac{dx}{x}\\\] now apply integral both sides
UsukiDoll
  • UsukiDoll
oh there it is the dv/dx has to be on the left and eventually that nasty problem turns into a simple separation of variables.
amoodarya
  • amoodarya
can you go on ?
amoodarya
  • amoodarya
it is one step to solve it completely!
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle -\sin(v){\tiny~~}dv=\frac{dx}{x} }\) and you integrate both sides... Just go, "magic", Blah Blah Blah?\ \(\color{#000000 }{ \displaystyle -\color{red}{\int}\sin(v){\tiny~~}dv=\color{red}{\int}\frac{dx}{x} }\) \(\tt Or,{\tiny~~~}are{\tiny~~~}you{\tiny~~~}integrating{\tiny~~~}with{\tiny~~~}respect{\tiny~~~}to{\tiny~~~}a{\tiny~~~}variable?\) Maybe something like this?\(\tiny\\[0.8em]\) \(\color{#000000 }{ \displaystyle -\sin(v){\tiny~~}dv=\frac{dx}{x} }\)\(\tiny\\[1.4em]\) \(\color{#000000 }{ \displaystyle -\sin(v){\tiny~~}\frac{dv}{dx}=\frac{1}{x} }\) and then, pam parara! \(\color{#000000 }{ \displaystyle -\color{red}{\int}\sin(v){\tiny~~}\frac{dv}{dx}\color{red}{dx}=\color{red}{\int}\frac{1}{x}\color{red}{dx} }\) I am just asking... Found this thread, but I know that this is a little wierd that I am replying to a 5-month-old thread.
UsukiDoll
  • UsukiDoll
yeahhhh....I don't understand how some users have the desire to post comments on an old thread but the first comment is the correct way of doing the problem. first substitution and then towards the end it becomes a simple (or hopefully...not that complex) separation of variables.

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