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Differential Equation

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider the differential equation:

    [tex] (x^2 + x) \frac {d^2y} {dx^2} - (x^2 - 2) \frac {dy} {dx} - (x + 2)y = 0[/tex]

    Given that [tex] y = e^x [/tex] is a solution, what is the general solution of this equation?

    2. Relevant equations

    ?


    3. The attempt at a solution

    I don't know where to start....please help :(
     
    Last edited: Nov 18, 2008
  2. jcsd
  3. Nov 18, 2008 #2

    gabbagabbahey

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    Since you are already given one solution, this would seem to suggest using the method of reduction of order....so assume an ansatz [itex]y_2(x)=v(x)y_1(x)=v(x)e^x[/itex] where [itex]v(x)[/itex] is an unkown function of [itex]x[/itex]...If you use the chain rule to find the first and second derivatives of [itex]y_2(x)[/itex] (they will involve [itex]v(x)[/itex] and its derivatives) and substitute them into your original DE, you can obtain a differential equation for [itex]v(x)[/itex] which should be easier to solve. Then just solve that DE and discard any solutions that make [itex]y_2(x)[/itex] linearly dependent on [itex]y_1(x)[/itex].
     
  4. Nov 18, 2008 #3
    So I have the derivatives (u = v(x)):

    [tex] y_2 = u y_1 = ue^x [/tex]

    [tex] y_2' = uy_1' + y_1u' = ue^x + e^xu' [/tex]

    [tex] y_2'' = uy_1'' + 2u'y_1' + y_1u'' = ue^x + 2u'e^x + e^xu'' [/tex]

    Putting these into the original DE...

    [tex] (x^2 + x) \frac {d^2y} {dx^2} - (x^2 - 2) \frac {dy} {dx} - (x + 2)y = 0[/tex]

    [tex] (x^2 + x)(ue^x + 2u'e^x + e^xu'') - (x^2 - 2)(ue^x + e^xu') - (x + 2)(ue^x) = 0[/tex]

    Not sure where to go with it next...my teacher gave a hint that there will be integration of parts.
     
  5. Nov 18, 2008 #4
    ok, well...to further things a little...

    I distributed everything out and added like terms, leading to:

    [tex] 3u'x^2e^x + 2u'xe^x - 2u'e^x + u''x^2e^x + u''xe^x = 0 [/tex]

    Using the variable transformation, [tex] w = u' [/tex] and [tex] w' = u'' [/tex]

    [tex] 3wx^2e^x + 2wxe^x - 2we^x + w'x^2e^x + w'xe^x = 0 [/tex]

    Still looks like a mess to me...?
     
  6. Nov 18, 2008 #5
    yea, I'm at a halt here..really not sure what to do next.
     
  7. Nov 18, 2008 #6

    gabbagabbahey

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    You might want to check your algebra again; I get:

    [tex] u'x^2e^x + 2u'xe^x + 2u'e^x + u''x^2e^x + u''xe^x = 0 [/tex]

    [tex] \Rightarrow e^x[u''(x^2+x) +u'(x^2+2x+2)] = 0 [/tex]

    [tex] \Rightarrow u''(x^2+x) +u'(x^2+2x+2)=0[/tex]

    [tex] \Rightarrow w'(x^2+x) + w(x^2+2x+2)=0[/tex]

    which, when you write [tex]w'=\frac{dw}{dx}[/tex], looks like a separable DE for w to me.....you know how to deal with those don't you? :wink:
     
  8. Nov 18, 2008 #7
    I redid the algebra by hand and got exactly what you got.

    Separating the variables

    [tex] -\frac {1}{w} dw = \frac {(x^2+2x+2)}{(x^2+x)}dx[/tex]

    integrating both sides and solving for w I get:

    [tex] w = \frac {((x+1) * e^-x)}{x^2} dx [/tex]

    So now u = integral of w dx
     
  9. Nov 18, 2008 #8

    gabbagabbahey

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    I assume you mean: [tex]w=\frac{(x+1)e^{-x}}{x^2}[/tex]?....if so, then good...now solve for [itex]u(x)[/itex] using your definition of w(x): [itex]w(x)=u'(x)[/itex]....then finally solve for [itex]y_2(x)=u(x)e^x[/itex]....what do you get?
     
  10. Nov 18, 2008 #9
    yea...i was editing the equation...still getting used to the code
     
  11. Nov 18, 2008 #10
    [tex] u(x) = -4 e^{-x} [/tex]

    so [tex] y_2(x) = -4 e^{-x} * e^x = -4 [/tex]

    correct?
     
  12. Nov 18, 2008 #11
    so would the general solution just be:

    [tex] y(x) = C_1 e^x + C_2 [/tex]
    Where -4 is included in the constant [tex] C_2 [/tex].
     
  13. Nov 18, 2008 #12

    gabbagabbahey

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    nope, I get [tex] u(x) = \frac{e^{-x}}{x} [/tex]

    give or take a constant multiplier.

    ...why don't you show me your work for integrate wdx?
     
  14. Nov 18, 2008 #13
    [tex] u(x) = \int w dx [/tex]

    [tex] u(x) = \int \frac {(x+1) e^-x}{x^2} dx = \int \frac {e^{-x}}{x} dx + \int \frac {e^{-x}}{x^2} dx[/tex]

    [tex] \int u dv = uv - \int v du [/tex]

    [tex] \int \frac {e^{-x}}{x} dx [/tex]

    [tex] u = \frac {1}{x} [/tex] [tex] du = -\frac {1}{x^2}dx [/tex]

    [tex] dv = e^{-x} dx [/tex] [tex] v = -e^{-x} [/tex]

    [tex] \int u v' = \frac {1}{x}(-e^{-x}) - \int -e^{-x} * - \frac {1}{x^2} dx = -2e^{-x}[/tex]

    Used same method for [tex] \int \frac e^{-x}}{x^2}dx [/tex] with different u du values, and got same answer, adding together got the [tex] -4e^{-x} [/tex]
     
  15. Nov 18, 2008 #14
    arg...made some typos on my calculator...not getting those answers anymore.
     
  16. Nov 18, 2008 #15

    gabbagabbahey

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    Do you mean:

    [tex]\int \frac {e^{-x}}{x} dx=\int uv'=\frac{-e^{-x}}{x}- \int \frac{e^{-x}}{x^2}dx[/tex]

    ....if so, there is no need to evaluate the integral on the right, since you are adding this to your other term:

    [tex]\Rightarrow u(x) = \int \frac {e^{-x}}{x} dx + \int \frac {e^{-x}}{x^2} dx=\left(\frac{-e^{-x}}{x}- \int \frac{e^{-x}}{x^2}dx \right)+ \int \frac {e^{-x}}{x^2} dx=\frac{-e^{-x}}{x}[/tex]

    ...I'm having a hard time seeing where your 4e^-x came from, but this is the way you should have done it.
     
  17. Nov 18, 2008 #16
    the -4e^-x came from using a minus sign instead of the negative key...
     
  18. Nov 18, 2008 #17

    gabbagabbahey

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    oh, lol....that's why I don't like using calculators :wink:
     
  19. Nov 18, 2008 #18
    been a while since I've done any integration by parts...

    ok...so with
    [tex] u(x) = - \frac {e^{-x}}{x} [/tex]

    [tex] y_2(x) = - \frac {e^{-x}}{x} * e^x = -\frac {1}{x} [/tex]

    Sooo...

    [tex] y(x) = C_1 e^x + C_2 (-x^{-1}) [/tex]
     
  20. Nov 18, 2008 #19
    that is correct, right? just checking, thanks for all the help!
     
  21. Nov 18, 2008 #20

    gabbagabbahey

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    Looks good to me :smile:

    ....although your prof might prefer that you make sure y1 and y2 are linearly independent by checking the Wronskian.
     
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