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More reduction of order DiffEq goodness!

  1. May 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Use the substitution [tex] y = (x^2 + 1)u [/tex] to solve the differential equation [tex] (x^2 +1)y\prime\prime = 2y [/tex]


    3. The attempt at a solution

    I was having some trouble with these earlier because I needed to brush up on my trigonometric substitution. Let's try this one...

    Making the substitution and simplifying the resultant equation gives us the differential equation [tex](x^2 + 1)u\prime\prime + 4xu\prime = 0[/tex]. Reducing the order then by substituting [tex] p = u\prime [/tex] [tex]p\prime = u\prime\prime[/tex] gives us the separable Deq [tex](x^2 +1)\frac{dp}{dx} + 4xp = 0 [/tex].

    So

    [tex] \frac{1}{p}dp = \frac{-4x}{x^2 +1} dx [/tex]

    [tex] ln|p| = -2 ln|x^2 + 1| + C_1 [/tex]

    [tex]p = C_1\frac{1}{(x^2 +1)^2} [/tex]

    substitute [tex] x = tan \theta[/tex] to integrate the above equation to get U

    [tex]u = C_1\int\frac{ sec^2\theta}{(1+tan^2\theta)^2}d\theta = C_1\int\frac{1}{sec^2\theta}d\theta = [/tex]

    [tex]C_1\int cos^2\theta d\theta [/tex]

    [tex]C_1\int \frac{1}{2} + \frac{cos2\theta}{2}d\theta = C_1(\frac{\theta}{2} + sin\theta cos\theta + C_2) [/tex]

    substituting back x for theta we get:

    [tex] u = C_1( \frac{tan^{-1} x}{2} + \frac{x}{x^2 +1} + C_2) [/tex]

    and substituting u into the first substitution:

    [tex] y = (x^2 +1)(C_1( \frac{tan^{-1} x}{2} + \frac{x}{x^2 +1}) + C_2). [/tex]


    Hopefully I've made fewer errors this time around! :wink:
     
    Last edited: May 24, 2010
  2. jcsd
  3. May 24, 2010 #2

    gabbagabbahey

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    [tex]\cos^2\theta\neq\frac{1}{2}[1+\sin(2\theta)][/tex]
     
  4. May 24, 2010 #3
    I don't see where I made that substitution? I now see I did make an error on this line:


    [tex]C_1\int \frac{1}{2} + \frac{cos2\theta}{2}d\theta = C_1(\frac{\theta}{2} + sin\theta cos\theta + C_2) [/tex]

    The integral of [tex]\frac{cos(2\theta)}{2} = \frac{sin(2\theta)}{4} [/tex], which I then replaced with a trigonometric identity should end up as [tex]\frac{sin(\theta)cos(\theta)}{2}[/tex]...fortunately in this case I don't think the damage is too great as the factor of 1/2 should be swallowed by the constant!
     
  5. May 24, 2010 #4

    gabbagabbahey

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    The factor of 1/2 isn't swallowed by the constant since the same constant multiplies another term. Fix the rest of your calculation with the 1/2 and you should be fine.
     
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