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anonymous

  • one year ago

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)

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  1. amoodarya
    • one year ago
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    take \[\frac{y}{x}=v\\y=vx\\y'=v+xv'\\(xcsc(\frac{y}{x})-y)dx+xdy=0\\(xcsc(\frac{y}{x})-y)+x\frac{dy}{dx}=0\\ divx\\(\csc(\frac{y}{x})-\frac{y}{x})+y'=0\\\]

  2. amoodarya
    • one year ago
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    if you put v it will be simple to solve can you go on ?

  3. UsukiDoll
    • one year ago
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    this is substitution method for first order odes

  4. anonymous
    • one year ago
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    i'll try :) our prof didn't really explained it to us very well the steps in solving it

  5. UsukiDoll
    • one year ago
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    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.

  6. anonymous
    • one year ago
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    actually i have already taken up integral and differential calculus but this is my first time encountering this topic

  7. UsukiDoll
    • one year ago
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    I really didn't like this method at all when I came across it x(

  8. UsukiDoll
    • one year ago
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    my favorites are integrating factor and exact oh yeah!

  9. amoodarya
    • one year ago
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    \[(\csc(v)-v)+(v+v'x)=0\\csc(v)+xv'=0\\xv'=-\csc(v)\\x\frac{dv}{dx}=\frac{1}{\sin(v)}\\-\sin(v)dv=\frac{dx}{x}\\\] now apply integral both sides

  10. UsukiDoll
    • one year ago
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    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.

  11. amoodarya
    • one year ago
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    can you go on ?

  12. amoodarya
    • one year ago
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    it is one step to solve it completely!

  13. SolomonZelman
    • one year ago
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    \(\color{#000000 }{ \displaystyle -\sin(v){\tiny~~}dv=\frac{dx}{x} }\) 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.

  14. UsukiDoll
    • one year ago
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    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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