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Homework Help: 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.

    when y(0)=1 and y'(0)=0

    3. The attempt at a solution
    This is my solution

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


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


    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


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    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 :)
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