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Solving 2nd order differential equation

  1. Dec 27, 2017 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    For the homogeneous equation, I have got the the root of the characteristic equation as ## e^{ix}, e^{-ix} ## .

    So, the corresponding solution is ## B \sin{ x} + A \cos{ x} ## .

    Then, I took the particular solution as C.

    Putting C in the differential equation , I get C = -1.

    So, the inhomogeneous solution is ## A \cos{ x} + B \sin{ x} – 1 ## , i.e. option(e).

    Is this correct?
  2. jcsd
  3. Dec 27, 2017 #2


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    Science Advisor
    Homework Helper
    Gold Member

    Looks great. You can plug it into the original differential equation to make sure that your solution satisfies it.
  4. Dec 27, 2017 #3


    Staff: Mentor

    You shouldn't need to ask this. Along the lines of @scottdave's advice, whenever you get a solution to a diff. equation, you should get in the habit of checking. That way you'll know whether your answer is correct. For an initial value problem, check that 1) your solution satisfies the initial condition, and 2) your solution satisifies the differential equation.
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