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Math Help - Sovling a differential equation

  1. #1
    Senior Member
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    Sovling a differential equation

    Hello folks.

    Let

    A = ]0,1[^2, \quad f(x) \in C^2(\overline{A})

    Find a solution f(x) for the following problem

    -f''(x) = const =: c > 0, \mbox{ iff } x \in A

    f(x) = 0, \mbox{ iff } x \in \partial A

    My try:

    -f(x) = \int \int c \ dx dx = \int c x + b dx = 0.5 c x^2 + bx + a

    f(0) = a = 0

    f(1) = 0.5 c + b = 0 \Rightarrow c = -2b

    So f(x) should be equal to 0.5 c x^2 -2b

    I'm not sure about this

    Rapha
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Rapha View Post
    Hello folks.

    Let

    A = ]0,1[^2, \quad f(x) \in C^2(\overline{A})

    Find a solution f(x) for the following problem

    -f''(x) = const =: c > 0, \mbox{ iff } x \in A

    f(x) = 0, \mbox{ iff } x \in \partial A

    My try:

    -f(x) = \int \int c \ dx dx = \int c x + b dx = 0.5 c x^2 + bx + a

    f(0) = a = 0

    f(1) = 0.5 c + b = 0 \Rightarrow c = -2b

    So f(x) should be equal to 0.5 c x^2 -2b

    I'm not sure about this

    Rapha
    You just proved that c= -2b so b= -\frac{1}{2}c.

    f(x)= \frac{1}{2}cx^2- \frac{1}{2}cx= \frac{1}{2}c(x^2- x)
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