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Math Help - Simple Initial Value Problem..

  1. #1
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    Unhappy Simple Initial Value Problem..

    hi guys,
    ok, so im reading my lecture notes again and here goes..

    Solve the IVP dx/dt = 2x^2 with condition x(5) = -3

    Solution: x(t) = -1/c + 2t
    At t=5, we require x(5) = -3 (not equal to 0). We can use the solution x(t) = -1/c+2t, and determine c from the data given, i.e. -3 = -1/c+2t
    so c = -29/3

    now this is where i dont understand! (how did they get 6t ?? )

    and hence the solution on the IVP is given by x(t) = 3/(29 - 6t)


    thanks guys!!
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  2. #2
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    Quote Originally Posted by matlabnoob View Post
    hi guys,
    ok, so im reading my lecture notes again and here goes..

    Solve the IVP dx/dt = 2x^2 with condition x(5) = -3

    Solution: x(t) = -1/c + 2t
    At t=5, we require x(5) = -3 (not equal to 0). We can use the solution x(t) = -1/c+2t, and determine c from the data given, i.e. -3 = -1/c+2t
    so c = -29/3

    now this is where i dont understand! (how did they get 6t ?? )

    and hence the solution on the IVP is given by x(t) = 3/(29 - 6t)


    thanks guys!!
    I think there's a problem. . When you spearate

    \frac{dx}{x^2} = 2 dt so - \frac{1}{x} = 2t + c or x = \frac{-1}{2t+c} , not what you have.
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  3. #3
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    aah of course ,how did i miss that?

    so its -1/c'+2t..

    im trying to get 6t somehow as the result does... is it possible to get 6t? =S because what im doing is plugging in c = -29/3 into x(t) and t=5 into x(t)... but im not getting any 6ts..
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  4. #4
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    Quote Originally Posted by Danny View Post
    I think there's a problem. . When you spearate

    \frac{dx}{x^2} = 2 dt so - \frac{1}{x} = 2t + c or x = \frac{-1}{2t+c} , not what you have.
    Now use the IC x(5) = -3 so -3 = \frac{-1}{10+c} giving c = - \frac{29}{3}.

    Therefore x = \frac{-1}{2t - \frac{29}{3}} = \frac{3}{29-6t}.
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  5. #5
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    Quote Originally Posted by matlabnoob View Post
    hi guys,
    ok, so im reading my lecture notes again and here goes..

    Solve the IVP dx/dt = 2x^2 with condition x(5) = -3

    Solution: x(t) = -1/c + 2t
    At t=5, we require x(5) = -3 (not equal to 0). We can use the solution x(t) = -1/c+2t, and determine c from the data given, i.e. -3 = -1/c+2t
    so c = -29/3

    now this is where i dont understand! (how did they get 6t ?? )

    and hence the solution on the IVP is given by x(t) = 3/(29 - 6t)


    thanks guys!!
    Alternative method:

    \frac{dx}{dt} = 2x^2

    \frac{dt}{dx} = \frac{1}{2x^2}

    \frac{dt}{dx} = \frac{1}{2}x^{-2}

    t = \int{\frac{1}{2}x^{-2}\,dx}

    t = -\frac{1}{2}x^{-1} + c

    \frac{1}{2x} = c - t

    2x = \frac{1}{c - t}

    x = \frac{1}{2c - 2t}.

    x = \frac{1}{C - 2t}, where C = 2c.


    Now by using the fact that x(5) = -3

    -3 = \frac{1}{C - 2(5)}

    -3 = \frac{1}{C - 10}

    -\frac{1}{3} = C - 10

    C = \frac{29}{3}.


    Therefore

    x = \frac{1}{\frac{29}{3} - 2t}

    x = \frac{3}{29 - 6t}.
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