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Math Help - need help please!

  1. #1
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    need help please!

    Solve the initial value problem dy/dx = (x 4)^4 where y(1) = 1.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bobby77
    Solve the initial value problem dy/dx = (x 4)^4 where y(1) = 1.
    The fundamental theorem of calculus tells us that all the functions
    whose derivative is (x-4)^4 are of the form:

    <br />
y(x)=\int_0^x (u-4)^4 du + C<br />

    So to solve this you need to do the integral, then plug in the initial condition
    to determin the value of constant that you require.

    RonL
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  3. #3
    MHF Contributor
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    Quote Originally Posted by bobby77
    Solve the initial value problem dy/dx = (x 4)^4 where y(1) = 1.
    "...where y(1) = 1"
    That means when x=1, then y=1.

    dy/dx = (x-4)^4
    y = (1/5)(x-4)^5 +C
    So, when x=1, and at the same time, y=1,
    1 = (1/5)(1 -4)^5 +C
    1 = (1/5)(-3)^5 +C
    1 = (1/5)(-243) +C
    Clear the fraction, multiply both sides by 5,
    5 = -243 +5C
    5 +243 = 5C
    C = 248/5

    Therefore,
    y = f(x) = (1/5)(x-4)^5 +248/5
    y = f(x) = (1/5)[(x-4)^5 +248] ------------answer.
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