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Higher order diff.eq undetermined coefficient

  1. Sep 11, 2015 #1
    What to do here if D=d/dx
    (D2+2D+4)y= x2e2x ?
    How to find particular integral by method of undetermined coefficient?
    if R.H.S would have been x2+ e2x then we could have taken
    yp= Ax2+ Bx+C + De2x
    but here in product, what to do?
     
  2. jcsd
  3. Sep 11, 2015 #2

    Geofleur

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    There may be a better way to do this but: Suppose you can solve ## (D^2 + 2D + 4)y = Cx^m ## for an arbitrary integer ## m ## and constant ## C ##. Then expand the exponential on the right hand side of the real problem using the power series representation ##e^{2x} = \sum_0^\infty \frac{2^n x^n}{n!} ##. The right hand side of the real problem will become a power series in ## x ##. Since you know how to solve ## (D^2 + 2D + 4)y = Cx^m ## for arbitrary ## C ## and ## m ##, you can solve the equation for each separate term of the series. Then the full (particular) solution is the sum of the solutions for each separate power of ## x ##.
     
  4. Sep 11, 2015 #3
    We have been not yet taught the power series. Any other method you know. Or apart from method of undetermined coefficients any other thing?
     
  5. Sep 11, 2015 #4
    The term on the right side of the equation is proportional to [itex]e^{2x} [/itex] so you should guess a particular solution that is proportional to [itex] e^{2x} [/itex]. So let [itex] y_p = f(x)e^{2x} [/itex] where [itex] f(x) [/itex] is a undermentioned function. Plug this expression for [itex] y_p [/itex] into the original differential equation. After applying the chain rule you'll be able to cancel at the factors of [itex]e^{2x} [/itex]. This will leave you with a differential equation for [itex] f(x) [/itex]. You only need to find one solution, and you should be able to do so using the method of undetermined coefficients.
     
  6. Sep 12, 2015 #5
    I'm arriving at f''(x)+4f'(x)+7f(x)=f(x)
    Is there also a method of inverse differential operator?
     
  7. Sep 12, 2015 #6
    You should double check your work. I get something different. For starters the right side of the equation should be x^2. Once you have an differential for f(x) you can solve it using what you have already learned. Remember that f(x) and y(x) are arbitrary symbols. You know how to solve an differential equation for y. The same methods are used to solve a differential equation for f.
     
  8. Sep 12, 2015 #7

    epenguin

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    An alternative - I multiply the equation through by e-2x.

    Then set a new variable Y = e-2xy

    Work out what Y' and Y" are.

    I got (D2 - 9) Y = x2

    I could have made a mistake, and if not it is not guaranteed beforehand or in other cases the LHS is quite so simple, but I think it is guaranteed you get the LHS still a 2nd order linear d.e. LHS in Y with constant coeffs and of course the RHS now of more well recognised type.
     
    Last edited: Sep 12, 2015
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