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    Me again... I'm doing an FP2 diff eq question, and I've come to a standstill.
    Here's the question:

    dx/dt + 6x/t = 2sqrt(x) e^(-t^2)
    Using the substitution z = x^(1/2) , solve the above differential equation.

    What I have:

    2z dz/dt + 6z^2/t = 2z e^(-t^2)

    dx/dt + 3z/t = e^(-t^2)

    t^3 dz/dt + 3z t^2 = t^3 e^(-t^2)

    d(z t^3)/dt = t^3 e^(-t^2)

    z t^3 = S t^3 e^(-t^2) dt

    Now, how on earth do I integrate  e^{-t^2} ???????? We haven't covered that at all!

    edit: **** latex
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    (Original post by StarvingAutist)
    Me again... I'm doing an FP2 diff eq question, and I've come to a standstill.
    Here's the question:
    

\frac{dx}{dt} + \frac{6x}{t} = 2\sqrt{x} e^{-t^2}

Using the substitution z = x^\frac{1}{2} , solve the above differential equation.

    What I have:

    

2z \frac{dz}{dt} + \frac{6z^2}{t} = 2z e^{-t^2}



\frac{dx}{dt} + \frac{3z}{t} = e^{-t^2}



t^3 \frac{dz}{dt} + 3z t^2 = t^3 e^{-t^2}



\frac{d(z t^3)}{dt} = t^3 e^{-t^2}



z t^3 = \int t^3 e^{-t^2} dt

    Now, how on earth do I integrate  e^{-t^2} ???????? We haven't covered that at all!
    Firstly, e^(-t^2) has no elementary antiderivative ie you can't integrate it. You can however integrate t^3e^{-t^{2}} .

    Spoiler:
    Show

    Consider the substitution u=t^2
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    (Original post by StarvingAutist)
    Me again... I'm doing an FP2 diff eq question, and I've come to a standstill.
    Here's the question:

    dx/dt + 6x/t = 2sqrt(x) e^(-t^2)
    Using the substitution z = x^(1/2) , solve the above differential equation.

    What I have:

    2z dz/dt + 6z^2/t = 2z e^(-t^2)

    dx/dt + 3z/t = e^(-t^2)

    t^3 dz/dt + 3z t^2 = t^3 e^(-t^2)

    d(z t^3)/dt = t^3 e^(-t^2)

    z t^3 = S t^3 e^(-t^2) dt

    Now, how on earth do I integrate  e^{-t^2} ???????? We haven't covered that at all!

    edit: **** latex
    You actually want \int e^{-t^2} t^3 dt, which is considerably simpler. Substitute u=t^2.
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    (Original post by ThatPerson)
    Firstly, e^(-t^2) has no elementary antiderivative ie you can't integrate it. You can however integrate t^3e^{-t^{2}} .

    Spoiler:
    Show

    Consider the substitution u=t^2
    (Original post by Smaug123)
    You actually want \int e^{-t^2} t^3 dt, which is considerably simpler. Substitute u=t^2.
    Thanks. I'm such a retard.
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    (Original post by StarvingAutist)
    Thanks. I'm such a retard.
    No problem - happens to us all!
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    IIRC, e^x^2 is a Gaussian Integral and it doesn't have an indefinite integral.
    That said, I never did Further Maths and I've only briefly covered Gaussian Integrals as part of my first term of a Chemistry degree.

    http://en.wikipedia.org/wiki/Gaussian_integral
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    (Original post by Smaug123)
    No problem - happens to us all!
    Some more than others :lol: :cry:
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    (Original post by TheWiseSalmon)
    IIRC, e^x^2 is a Gaussian Integral and it doesn't have an indefinite integral.
    That said, I never did Further Maths and I've only briefly covered Gaussian Integrals as part of my first term of a Chemistry degree.

    http://en.wikipedia.org/wiki/Gaussian_integral
    Yeah, the error function is what came up on wolfram alpha.
 
 
 
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