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How to solve this DE

  1. May 1, 2010 #1
    I have to find three ways of solving the equation

    x*y'' - y' = 3x^2

    So far, I have tried Cauchy-Euler with success and am now working on using series solutions (although I do not know a series for 3x^2). And for a final way of solving I am completely lost.
     
  2. jcsd
  3. May 2, 2010 #2

    Char. Limit

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    Gold Member

    Is there any way that you could "un-product rule" the left side into d/dx(xy') or something? That might work... or it might not.

    You could always try the "Professor's Solution"...

    "This problem is left as an exercise for the grader."
     
  4. May 2, 2010 #3
    This problem is solved with 'method of variation of parameters'.this way,y''+p(x)y'+q(x)y=f(x)
    Yp=c1*y1+c2*y2
    Yp(x)=u1*y1+u2*y2
    u1=(-)integration(y2*f(x)/w[y1,y2])dx and u2=int(y1*f(x)/w[y1,y2])dx
    w[y1,y2] is wronskian of y1 and y2
    consequently,we locate yg=yp+yc
     
  5. May 2, 2010 #4
    You can either let u=y' or just multiply both side by 1/x^2 and get [itex]d\left(\frac{1}{x}y'\right)=3[/itex]
     
  6. May 2, 2010 #5
    I have used the cauchy-euler with variation of parameters with success. I have been informed by a friend and now jackmell to use the substitution u=y'. unfortunately i do not know how that is worked out. if i let u=y' then the order of the equation is reduced. does that mean i can just solve it as a first order linear equation? or are there other steps i need to be aware of.

    Thanks everyone for the input! This site is awesome
     
  7. May 2, 2010 #6
    I should have written [itex]d\left(\frac{1}{x}y'\right)=3dx[/itex]

    to be more accurate but that's just skipping steps so yes, make the substitution u=y', solve the first-order equation in u for u, then back-substitute dy/dx=u, and solve for dy, then integrate again. Don't forget to include the constants of integration after each integration.
     
  8. May 2, 2010 #7
    Thanks...i did so and got an answer! So far, i have completed two methods successfully yet they both produce a slightly different answer. I have yet to plug them into the original but Im curious, does this happen when various methods are used?
     
  9. May 2, 2010 #8
    You mean the "same" answer just written differently. For example if c is an arbitrary constant then [itex]e^c x^2[/itex] and [itex]c x^2[/itex] represent the "same" answer. Yes, that happens especially when you use Mathematica to check your work but you can just do a little algebra to get it to look like the other expression or in the above, recognize that [itex]e^c[/itex] is still just an arbitrary constant. Just type:

    DSolve[x y''[x]-y'[x]==3x^2,y,x]

    into Wolfram Alpha to check your result.
     
  10. May 2, 2010 #9
    What is mathematica?
     
  11. May 2, 2010 #10
    Mathematica is a program for doing mathematics. You can google Wolfram Alpha which accesses some of the features of Mathematica. You typed in that DSolve command yet? Well, when you do, it will pass it to a running Mathematica program on some server somewhere and it will solve the differential equation and give you the result. Of course don't rely on asking Mathematica to solve problems for you. I recommend it here to check your answers that you first learn to work manually--then use Mathematica :)
     
  12. May 3, 2010 #11
    Power series for 3x^2 is 3x^2 (all other terms are zero).
     
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