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Differential Equation. A little help please kind lads

  1. Dec 13, 2005 #1
    Differential Equation. A little help please kind lads :)

    Hey, ive got this problem... i pay attention in class and read the chapter in the book but i cant seem to know how to solve this? ANy help is greatly appreciated lads... thank you

    y'tanx=a+y y(pi/3)=a 0<x<(pi/2)

    What i done

    (dy/dx)tanx = a + y
    dytanx=(a+y)dx
    dy/(a+y)=dx/tanx

    integral of both sides...

    ln(a+y)=-csc^2(x) ??? This isnt the answer in the back of the book, anyone can tell me where ive gone wrong? 1/tanx = cot x and the integeral of cotan is -csc^2.... oh i also assumed a was a constant, is it not? :S :( thanks guys
     
  2. jcsd
  3. Dec 13, 2005 #2
    Consider an alternative method of integrating cot(x), try doing a u substitution with u=sin(x).
     
  4. Dec 13, 2005 #3
    LN(a+y) = LN (sinx) + C

    Like this?

    Then

    a+y = sinx +c (take e^ of both sides)

    so y = sinx - a

    the answer is y= (4a*sinx)/sqrt(3) - a

    my answer is close to that i suppose now i have to do something with that y(p/3) = a ... any hints?
     
  5. Dec 13, 2005 #4
    sorry for the double post ... my internet messed up..


    im tryin to see how to slve y(pi/3) = a now
     
  6. Dec 13, 2005 #5
    You did the integration correctly, however there's an algebraic mistake when you raised both sides of your equation to the power of e:

    [tex]e^{\ln(\sin{x}) + c} \neq \sin{x} + c[/tex]
    [tex]e^{\ln(\sin{x}) + c} = e^{\ln(\sin{x})} e^{c} = c \sin{x}[/tex]

    (relabling e^c as just c, since they're both constants anyway).

    Also, when you're matching boundary conditions the general method of attack is to evaluate the relevant function at the value specified then set it equal to the value of the boundary condition.
     
  7. Dec 13, 2005 #6
    so i assume that i did a mistake on the a+y side as well?
     
  8. Dec 13, 2005 #7
    No, that side's fine since the a+y is all within the natural log.
     
  9. Dec 13, 2005 #8
    ok so i have

    y=e^c sinx - a

    now i have to do y(pi/3) = e^c*sin(pi/3)-a =a and solve for c?

    *try*
     
  10. Dec 13, 2005 #9
    thanks a lot i got it now, i really appreciate it :) thank you so much
     
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