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Homework Help: CLEP Test Question

  1. Nov 4, 2009 #1
    1. The problem statement, all variables and given/known data

    I was looking into taking the Calculus CLEP Exam and came across this practice problem:

    Derive the general solution to the following differential equation: y' = sin x.

    A. sin y + cos x = C
    B. tan y = C
    C. sin y - cos x = C
    D. cos x = C
    E. cos x - sin y = C

    3. The attempt at a solution

    The first thing I thought was that the question was asking for me to integrate y'=sin X. If y' = sin X, then y = -cos x + C. I didn't know what they wanted me to do so I simply set this equal to zero, so -cos x + C = 0 and therefore cos X=C. However, D is not the correct answer. So, my question is: what is it that the question is asking me to do when it says it wants a "general solution to ...[a] differential equation"?
  2. jcsd
  3. Nov 4, 2009 #2
    It might be relevant to state that choices A-E are answers for a multiple choice problem and not five parts to the problem.
  4. Nov 4, 2009 #3


    Staff: Mentor

    Your first inclination was correct. The solution to this differential equation is y = -cos x + C. That is the general solution. I have no idea why this is not listed as an option. Is what you showed the exact wording of the problem?
  5. Nov 4, 2009 #4
    Yea, the text I provided is the exact wording. The answer, which was listed at the bottom of the page, was listed as (A) and states:

    A. In order to be correct, a general solution to a differential equation must satisfy both the homogenous and non-homogenous equations.
  6. Nov 4, 2009 #5


    Staff: Mentor

    Our solution satisfies the DE. If y = -cos x + C, then dy/dx = sin x.

    The homogeneous problem would be y' = 0, for which the solution is y = C. The nonhomogeneous problem is y' = sin x, for which the general solution is y = -cos x + C.

    I think there might be a typo in the CLEP answer. They might have meant to say y + cos x = C, which would be equivalent to our answer, instead of sin y + cos x = C.
  7. Nov 5, 2009 #6
    Can you explain what the homogeneous and nonhomogeneous problems are? I haven't come across those terms before.
  8. Nov 5, 2009 #7


    Staff: Mentor

    A differential equation can be represented as f(y, y', y'', ..., y(n)) = g(x). On the left side you have some function of y and its derivatives. On the right side there is some other function of the independent variable.

    The homogeneous equation is f(y, y', y'', ..., y(n)) = 0. The one above is the nonhomogeneous equation. For your problem, you have y' = sin x. (There is no y term.) The homogeneous equation is y' = 0, and the nonhomogeneous equation is y' = sin x.

    Another example is y'' + y = x, a nonhomogeneous DE. The related homogeneous equation is y'' + y = 0, whose solution is y = Asin x + Bcos x. The general solution of the nonhomogeneous equation turns out to be y = Asin x + Bcos x + x, and is made up of the solution to the homogeneous equation plus what is called a particular solution to the nonhomogeneous problem.

    The term "homogeneous" is also used in a different context in differential equations, which can lead to some confusion. In that sense of the word, a differential equation in the form y' = f(y/x) is said to be homogeneous. My sense is that in the world of differential equations, the description I gave before is much more commonly used.
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