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Solve general differential equation: please check my work?

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data
    I forgot how to do these and just need a little refreshing on how to solve these


    2. Relevant equations

    y' - 2y = (t^2)(e^2t)


    3. The attempt at a solution

    y' - 2y = (t^2)(e^2t)

    ok so my integrating factor is the value in front of y, right?
    so mu(x) = e^int-2
    = e^-2t

    so then after i multiply whole equation by e^-2t i get:
    (e^-2t)y = t^2 (because the e^-2t cancelled the e^2t that was in the right hand term).


    SO this is where I totally forgot how to proceed!! Do i integrate both sides, then im done? or am i missing a step?

    and if i do integrate then do i use u-sub for the term on LHS?
    thanks for any help


    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 8, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    Remember your LHS always becoms

    d/dt( y*integratingn factor)


    so now you just integrate both sides.
     
  4. Apr 11, 2010 #3
    integ (ye^-2t)' = integ t^2

    = ye^-2t = (1/3)t^3

    y = (t^3) / (3e^-2t) + c (this is my final solution)

    Is this correct? Also, even though I dont have initial cond can i say anything more about what c is?

    thanks
     
  5. Apr 11, 2010 #4

    rock.freak667

    User Avatar
    Homework Helper

    The line in red should be

    ye-2t+(1/3)t3+C

    then
    you can divide by e-2t
     
  6. Apr 11, 2010 #5
    thanks for the reply..but i dont understand what and why you did that? what does that equal? If zero, then how did (1/3)t^3 become negative?
     
  7. Apr 11, 2010 #6

    rock.freak667

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    Homework Helper

    Terribly sorry, I made a typo, it should be this.

    ye-2t=(1/3)t3+C
     
  8. Apr 11, 2010 #7
    ok thanks. so only diff was that i need to reference c right after integration before doing anything else to equation, --right?
     
  9. Apr 11, 2010 #8

    Mark44

    Staff: Mentor

    In the line above, the first = shouldn't be there, and you should add the constant of integration in this step.

    Each line should be an equation that implies, but is not equal to, the next line. After all, an equation can't be equal to another equation.

    To recap,
    [tex]d/dt(ye^{-2t} = t^2[/tex]
    [tex]\Rightarrow ye^{-2t} = \int t^2 dt = \frac{t^3}{3} + C[/tex]
    [tex]\Rightarrow y = e^{2t}(\frac{t^3}{3} + C)}[/tex]

    Notice also that the constant is multiplied by e2t.

     
  10. Apr 12, 2010 #9
    thanks for the replies..
    so my final solution is:

    y = (1/3) (t^3)((e^t)^2) + ce^t^2

    is this correct? also, since there were no init cond, is this whats called the general solution to the DE? If i did have IC i would be able to know what c is right?
    thanks
     
  11. Apr 12, 2010 #10

    Mark44

    Staff: Mentor

    No, not correct. See post #8. Where are you getting et2? And why are you writing e2t as (et)2?

    Yes, if you had an initial condition you could solve for C.
     
  12. Apr 12, 2010 #11
    should be:

    y = (1/3)(e^2t)(t^3) + c(e^2t) (i integrated integrating factor incorrectly)

    is this correct? (also since there were no init cond, is this whats called the general solution to the DE? If i did have IC i would be able to know what c is right?
    thanks)
     
  13. Apr 12, 2010 #12

    Mark44

    Staff: Mentor

    Why are you asking this? Look at post #8.
    Yes, this is the general solution. The other question has already been answered.
     
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