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General solution of differential equation (express y in term of x)

  1. May 3, 2014 #1
    1. The problem statement, all variables and given/known data

    i got stucked here. below is the answer given. can anybody help please?

    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. May 3, 2014 #2

    haruspex

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    You can try to simplify the original equation by substituting y(x) = f(x).xn, then see what value of n will get rid of the -y/x term.
     
  4. May 3, 2014 #3
    This looks like a straightforward "integrating factor" problem.

    Chet
     
  5. May 3, 2014 #4
    dy/dx +Py(x) = Q(X) if i rearrange i would get 0.5 dy/dx + y/x = arc tan x ... is arc tan x function of x?
     
    Last edited: May 3, 2014
  6. May 3, 2014 #5
    Multiply both sides by 2. arc tan x is a function of x.

    Chet
     
  7. May 4, 2014 #6
    i redo the question and dont know how to proceed here... any idea on how should i do next ? i dont know how to integreate arc tan x
     

    Attached Files:

  8. May 4, 2014 #7
    Integrate by parts. Do you remember how to take the derivative of arc tan x with respect to x?

    Chet
     
  9. May 4, 2014 #8
    sorry , i checked thru the syllabus, there's no deriative of arc tan x in it or maybe i can do it in other way? any other way?
     
  10. May 4, 2014 #9
    Yes. Let y = arc tan x

    Then tan y = x

    Differentiating both sides with respect to x;

    [tex]sec^2y\frac{dy}{dx}=1[/tex]

    Also, we have the trig identity: [itex]tan^2y+1=sec^2y[/itex]

    So, [itex]sec^2y=1+x^2[/itex]

    So, [tex]\frac{dy}{dx}=\frac{1}{1+x^2}[/tex]

    So, [tex]\frac{d(tan^{-1}x)}{dx}=\frac{1}{1+x^2}[/tex]

    Chet
     
  11. May 4, 2014 #10
    thanks got the solution finally!
     
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