1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Have I done this differential equation correctly?

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data
    This is my first big differential equation that I have solved by myself, and I wanted to see if it is correct done, or what I have done wrong with it.
    Would really appreciate if somebody could take a look on it.

    y''+6y'+8y=0
    when y(0)=1 and y'(0)=0



    3. The attempt at a solution
    This is my solution
    y''+6y'+8y=0
    r2+6r+8=0

    r=-3+- sqrt (9-8)= sqrt 1
    r1= -2
    r2= -4

    y=Ce-2x+De-4x
    y'=-2Ce-2x-4De-4x

    y(0)=1 1=Ce0+De0=C+D
    y'(0)=0 0=-2Ce0-4De0=-2C-4D

    1=C+D
    0=-2C-4D

    I take 1=C+D and take -D on both sides so I get 1-D=C

    I take 0=-2C-4D and take +2 on both sides so I get 2=C-4D
    I now put in 1-D in it, and get 2=1-D-4D and that I can change to 3=-5D
    I then take +5 on both sides and i get that 8=D

    I go back to 1-D=C and I get that C=-7

    I now take C and D and put them in the equation, so I get:

    -7e-2x+8e-4x

    Is this right, or have I made something wrong on the way?
     
  2. jcsd
  3. Sep 1, 2012 #2
    Re: Have I done this differential equation correct?

    This is a question you can answer yourself. Just check that you solution satisfies the equation and the initial conditions.
     
  4. Sep 1, 2012 #3
    Re: Have I done this differential equation correct?

    I don't wanna come across as stupid, but how do I do that?
     
  5. Sep 1, 2012 #4
    Re: Have I done this differential equation correct?

    You have found y(x). Let x = 0, what do you get? It should be 1 as per initial conditions. Then differentiate your y(x). What is y'(0)? It should be 0 as per the conditions. Finally, differentiate that again, and plug your y, y' and y'' in the equation. Is it satisfied?
     
  6. Sep 1, 2012 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Also, since solving for C and D was a little subproblem, you should also check if your C and D satisfy the simultaneous equations. You should get in the habit of sanity-checking your work as you go.
     
  7. Sep 1, 2012 #6
    I think it gets wrong after what I highlighted.
    Is there someone who can explain a little bit, so I can get in the right direction of solving this one?
     
  8. Sep 1, 2012 #7
    You had 0 = -2C - 4D.

    This somehow became 2 = C - 4D.

    It is you who should explain how this is ever possible. You said "I take +2 on both sides", but that can't be it. +2 on both sides would get you 2 = -2C - 4D + 2
     
  9. Sep 1, 2012 #8
    i see it now, I thought in the wrong way, I cant take the 2 away from C.
    Can I do it like this?

    2*1 + 0 = 2*(C+D) - 2C - 4D ⇔ 2 = 2C - 2C + 2D - 4D ⇔ -2D = 2 ⇔ D = -1

    This gives 1 = C - 1 ⇔ C = 2
     
  10. Sep 2, 2012 #9
    This is better. Now, check that your result satisfies the conditions of the problem.
     
  11. Sep 2, 2012 #10
    Everything adds up, it must be correct!
    What a great feeling I get from seeing that it is right :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Have I done this differential equation correctly?
  1. What have I done? (Replies: 1)

Loading...