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Impossible for me to integrate??

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−t2 e^{-t^2} ???????? We haven't covered that at all!

edit: **** latex
(edited 9 years ago)
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:
[br]dxdt+6xt=2xe−t2[br]Usingthesubstitutionz=x12,solvetheabovedifferentialequation.[br][br]\frac{dx}{dt} + \frac{6x}{t} = 2\sqrt{x} e^{-t^2}[br]Using the substitution z = x^\frac{1}{2} , solve the above differential equation.[br]

What I have:

[br]2zdzdt+6z2t=2ze−t2[br][br]dxdt+3zt=e−t2[br][br]t3dzdt+3zt2=t3e−t2[br][br]d(zt3)dt=t3e−t2[br][br]zt3=∫t3e−t2dt[br][br][br]2z \frac{dz}{dt} + \frac{6z^2}{t} = 2z e^{-t^2}[br][br]\frac{dx}{dt} + \frac{3z}{t} = e^{-t^2}[br][br]t^3 \frac{dz}{dt} + 3z t^2 = t^3 e^{-t^2}[br][br]\frac{d(z t^3)}{dt} = t^3 e^{-t^2}[br][br]z t^3 = \int t^3 e^{-t^2} dt[br][br]

Now, how on earth do I integrate e−t2 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 t3e−t2t^3e^{-t^{2}} .

Spoiler

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−t2 e^{-t^2} ???????? We haven't covered that at all!

edit: **** latex

You actually want ∫e−t2t3dt\int e^{-t^2} t^3 dt, which is considerably simpler. Substitute u=t2u=t^2.
Original post by ThatPerson
Firstly, e^(-t^2) has no elementary antiderivative ie you can't integrate it. You can however integrate t3e−t2t^3e^{-t^{2}} .

Spoiler



Original post by Smaug123
You actually want ∫e−t2t3dt\int e^{-t^2} t^3 dt, which is considerably simpler. Substitute u=t2u=t^2.


Thanks. I'm such a retard. :frown:
Original post by StarvingAutist
Thanks. I'm such a retard. :frown:

No problem - happens to us all!
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
Original post by Smaug123
No problem - happens to us all!


Some more than others :lol: :cry:
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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