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Find a linear homogeneous equation with given general solution

  1. May 15, 2013 #1
    I need help finding a linear homogenous constant-coefficient differential equation with the given general solution.


    2. I tried to come with differential equation but this is it
    I can 't seem how to begin
  2. jcsd
  3. May 15, 2013 #2
    I came with taking the fourt derivate of y''''=Ce^-x(C3(x2-8x+12)+C2(x-4)+C+e^2x so Idont know what to do next. help please asap.
  4. May 15, 2013 #3
    is the the differential equation -y''''-y''=0 im not sure
  5. May 15, 2013 #4
    The general solution implies that the characteristic equation has one distinct real root and one repeated real root, it can be factored as (r-1)(r+1)^3
  6. May 16, 2013 #5
    so if its factored can be written as (r-1)r^3

    and alternate fom beign r^4-r^3

    making the general solution


    is this alright thanks.
  7. May 16, 2013 #6
    Not quite. If you multiply (r-1)(r+1)[itex]^{3}[/itex], you should get
  8. May 16, 2013 #7
    making the differential equation


    is this alright and sorry about my mistake in factoring
  9. May 16, 2013 #8
    The differential equation would be


    Remember that r=r[itex]^{1}[/itex] and 1=r[itex]^{0}[/itex]
  10. May 16, 2013 #9
    thanks can help solve this problem

    I have a problem which in involves a second order differential equations with imaginary roots and I can seem to know how to finish the problem.

    d^2y/dt^2 +15y =cos 4t+2 sin t

    this is what I got so far

    r^2+15=0 for the homogeneous part



    and now the part that follows is to come with a particular solution for

    cos 4t+2 sin t

    but I don't know how to properlly set up
  11. May 16, 2013 #10
    and thanks for helping me so far
  12. May 16, 2013 #11
    I know that 2 is a constant but can my particular solution be

    Yh=Acos4t +Bsin4t
  13. May 16, 2013 #12


    Staff: Mentor

    Problems such as these should be posted in the Homework & Coursework section, not in the technical math sections. I am closing this thread.
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