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Am i doing this right

  1. Feb 8, 2006 #1
    Hey, just wanna check iv done these questions right so far...

    [tex] \mbox{Find the general solution to} \frac{dy}{dx} = y^2xcos(2x) \mbox{giving explicitly in terms of x.} [/tex]
    [tex]\mbox{Find the particular solution satisfying y(0) = -1} [/tex]

    My answer:

    [tex] \frac{dy}{y^2} = x\cos(2x)dx \Rightarrow \int\frac{dy}{y^2} = \intx\cos(2x)dx [/tex]
    [tex]u = x [/tex]
    [tex]\frac{du}{dx} = 1 [/tex]
    [tex]\frac{dv}{dx} = \cos2x [/tex]
    [tex]v = \frac{1}{2}\sin(2x)} [/tex]
    [tex]\Rightarrow \frac{xsinx}{2} - \frac{1}{2}\int\sin(2x) = \frac{x\sinx}{2} + \frac{1}{4}\cos(2x) + C[/tex]

    So we have

    [tex] \frac{-1}{y} = \frac{x\sin(x)}{2} + \frac{1}{4}\cos(2x) + C[/tex]

    So

    [tex] y = \frac{-4}{\cos(2x)} - \frac{2}{x\sin(2x)} + C [/tex]

    [tex]\mbox{at y(0) = -1 \Rightarrow 1 = \frac{-4}{1} - \frac{2}{0} + C \Rightarrow C = 5}[/tex]

    So overall,

    [tex]y = \frac{-2}{x\sin(2x)} - \frac{4}{\cos(2x)} + 5}. [/tex]
     
    Last edited by a moderator: Feb 8, 2006
  2. jcsd
  3. Feb 8, 2006 #2

    Tom Mattson

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    Staff Emeritus
    Science Advisor
    Gold Member

    OK.


    Nope. You can't take the reciprocal of a fraction that way. If you could then we would have the following:

    [itex]\frac{1}{2}=\frac{1}{3}+\frac{1}{6}[/itex] (True)

    So

    [itex]2=3+6[/itex] (False)

    Instead you must combine the two terms on the right with a common denominator, and then take the reciprocal.
     
  4. Feb 8, 2006 #3
    [tex] \mbox{2. Find a particular integral for each of the following equations:}[/tex]

    [tex]/mbox{ i. \frac{d\theta}{dz} + 2\theta = 8} [/tex]
    [tex]/mbox{ ii. \frac{dx}{dt} - 2x = 14e^{-5t}} [/tex]
    [tex]/mbox{iii. \frac{dx}{dt} + x - -3sin2t + 4cos2t} [/tex]

    My answers:

    [tex]\mbox{ i. Constant term so choose x = a + bt} [/tex]
    [tex] \theta(z) = 1 \Rightarrow \frac{d\theta}{dz} = 0 \Rightarrow 0 + 2a = 8 \rightarrow a = 4 \Rightarrow PI = \Theta(z) = 4. [/tex]

    [tex]\mbox{ ii. Choose x = ae^(-t). This gives -5ae^{-5t} - 2ae^{-5t} = 14e^{-5t} \Rightarrow a = -2 \Rightarrow PI: x(t) = -2e^{-5t)}[/tex]

    [tex]\mbox{ iii. All i know to do here so far is choose x = acos2t + bsin2t. How do i continue?}[/tex]
     
  5. Feb 8, 2006 #4
    Sorry il sort the coding to this when i have a second
     
  6. Feb 13, 2006 #5
    That post shoul have read:

    [tex] \mbox{2. Find a particular integral for each of the following equations:}[/tex]

    [tex] i. \frac{d\theta}{dz} + 2\theta = 8 [/tex]
    [tex] ii. \frac{dx}{dt} - 2x = 14e^{-5t} [/tex]
    [tex]iii. \frac{dx}{dt} + x - -3sin2t + 4cos2t [/tex]

    My answers:

    [tex] i. Constant term so choose x = a + bt [/tex]
    [tex] \theta(z) = 1 \Rightarrow \frac{d\theta}{dz} = 0 \Rightarrow 0 + 2a = 8 \rightarrow a = 4 \Rightarrow PI = \Theta(z) = 4. [/tex]

    [tex] ii. Choose x = ae^(-t). This gives -5ae^{-5t} - 2ae^{-5t} = 14e^{-5t} \Rightarrow a = -2 \Rightarrow PI: x(t) = -2e^{-5t)[/tex]

    iii. All i know to do here so far is choose x = acos2t + bsin2t. How do i continue?
     
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