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Two-point boundary value problem

  1. Nov 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve the given BVP or show that it has no solution. (It does have a solution)
    y"+2y = x, y(0)=y([itex]\pi[/itex])=0


    2. Relevant equations
    Characteristic polynomial is r^2 + 2 = 0. μ = √2



    3. The attempt at a solution
    The solution to the complementary homogeneous equation is y_h = c1 cos(√2x) + c2 sin(√2x)
    Since the BVP is not homogeneous, there is a solution for the nonhomogeneous part. Let's call it y_c = d1*x + d2. Upon substituting into the problem, d1=1/2 and d2=0.

    The solution is of the form y = c1 cos(√2x) + c2 sin(√2x) + (1/2)x

    This was the way a similar problem was solved in the textbook. Same boundary conditions but the eqn was y"+y=x instead of y"+2y=x

    The solution on the back is of the form y = c1*sin(√2x) + c2*x*sin(√2x).
    Why is that?
     
  2. jcsd
  3. Nov 30, 2012 #2

    haruspex

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    Misprint? You can easily check that this is not a solution of the equation given.
     
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