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Initial value problem

  1. Jul 19, 2014 #1
    For the following problem

    [itex]\frac{dw}{d\theta}=\theta w^{2}sin(\theta^{2}), w(0)=1[/itex]

    I am not able to obtain the solution

    [itex]w=\frac{2}{1+cos(\theta^{2})}[/itex]

    Can anyone point out my mistake?

    I have attached my working out in a picture format below (may need to enlarge it)

    thanks
     

    Attached Files:

  2. jcsd
  3. Jul 19, 2014 #2

    SteamKing

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    sin(θ[itex]^{2}[/itex]) ≠ sin[itex]^{2}[/itex](θ)

    Be careful where you put the exponents when using trig functions. It makes a difference for integrating and differentiating as well.
     
  4. Jul 20, 2014 #3
    sorry wrong piece of paper.

    Please look at this new attachment and inform me of my error
     

    Attached Files:

  5. Jul 20, 2014 #4

    verty

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    You forgot the arbitrary constant.
     
  6. Aug 21, 2014 #5

    rjr

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    nothing wrong with the u-substitution and integration, however on the right side toward the bottom of your page, one step reads:

    1/w = 2/(cos(Θ2)) + C

    What I suggest:

    1). ALWAYS isolate the function variable first (in this case w(Θ)) before solving for the arbitrary constant (in this case C).

    In other words - rather than what it reads now: 1/w(Θ) = ...

    rearrange the equation such that it reads w(Θ) = ...

    THEN solve with the initial value w(0) = 1

    2). AlSO (and more importantly) keep in mind that any value multiplied by an arbitrary constant is still the value of the arbitrary constant (e.g. 2 * C = C)

    So putting these two suggestions together, rewrite 1/w = 2/(cos(Θ2)) + C such that it reads 1/w(Θ) = .... then one ENTIRE fraction. Then flip the fractions on either side so that it reads w(Θ) = ...

    THEN solve with the initial condition w(0) = 1.

    hope this helps
     
    Last edited: Aug 21, 2014
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