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Math Help - -u(x)'' = -pi^2*sin(pi*x) ?

  1. #1
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    -u(x)'' = -pi^2*sin(pi*x) ?

    Hi, whats the solution to

    -u(x)'' = -pi^2*sin(pi*x)

    subject to boundary conditions u(0) = 0, u'(1) = 0

    in terms of u(x) = ...

    i get u(x) = sin(pi*x) - pi*x , but i think its very wrong

    any help much appreiciated thanks !
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  2. #2
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    Quote Originally Posted by Richyie View Post
    Hi, whats the solution to

    -u(x)'' = -pi^2*sin(pi*x)

    subject to boundary conditions u(0) = 0, u'(1) = 0

    in terms of u(x) = ...

    i get u(x) = sin(pi*x) - pi*x , but i think its very wrong

    any help much appreiciated thanks !
    Not quite right. Certainly that u satisfies the differential equation and u(0)= 0- pi(0)= 0. But u'= pi cos(pi x)- pi so u'(1)= pi cos(pi)- pi and, since cos(pi)= -1, u'(1)= -pi- pi= -2pi, not 0. But that's easy to fix!
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