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Math Help - Integrating IVP

  1. #1
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    Thumbs down Integrating IVP

    could someone please please tell me how to integrate

    dx/dt = (x^3).(e^t)

    intial condition : x(1) = -4

    this looks v.simple but trust me, its not!

    help me out! aaaah...

    thank you
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  2. #2
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    Quote Originally Posted by matlabnoob View Post
    could someone please please tell me how to integrate

    dx/dt = (x^3).(e^t)

    intial condition : x(1) = -4

    this looks v.simple but trust me, its not!

    help me out! aaaah...

    thank you

    \frac{dx}{dt}=x^3e^t

    \frac{dx}{x^3}=e^tdt

    Now integrate both sides

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

    can you handle it from here by plugging in your intial value to find c
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  3. #3
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    Quote Originally Posted by artvandalay11 View Post
    \frac{dx}{dt}=x^3e^t

    \frac{dx}{x^3}=e^tdt

    Now integrate both sides

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

    can you handle it from here by plugging in your intial value to find c

    thank you!

    i have gotten that far and then i ended up with =S...
    after i substituted 1 into t..

    x^2 = -1/2e+2c

    ?

    that looks v.v.wrong to me. or is it just me =S
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  4. #4
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    Quote Originally Posted by matlabnoob View Post
    thank you!

    i have gotten that far and then i ended up with =S...
    after i substituted 1 into t..

    x^2 = -1/2e+2c

    ?

    that looks v.v.wrong to me. or is it just me =S

    x(1)=-4 so when t=1, x=-4

     <br />
-\frac{1}{2(-4)^2}=e^1+c<br />

    -\frac{1}{32}=e+c

    So c=-e-\frac{1}{32}

    And the solution is

    -\frac{1}{2x^2}=e^t-e-\frac{1}{32}

    which can be combined or manipulated or whatever you wanna do but I'll leave it like that
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  5. #5
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    thank you!=] woaah its that easy.... i spent 2 hourson this!!
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