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 Post subject: separation of variables involving a quadratic
PostPosted: Tue, 7 Aug 2012 21:53:01 UTC 
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given problem:

dx/dt=k(3-x)(2-x)

x(0)=1

I have a worked solution for this problem and it uses partial fractions as a key method.

My question is whether this problem can be solved without partial fractions, ie: combining terms to produce a quadratic in x.

A yes or no is what I am looking for. If yes I will attempt to work the problem.


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 Post subject: Re: separation of variables involving a quadratic
PostPosted: Wed, 8 Aug 2012 03:09:44 UTC 
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canadianicon25 wrote:
given problem:

dx/dt=k(3-x)(2-x)

x(0)=1

I have a worked solution for this problem and it uses partial fractions as a key method.

My question is whether this problem can be solved without partial fractions, ie: combining terms to produce a quadratic in x.

A yes or no is what I am looking for. If yes I will attempt to work the problem.


Recall \displaystyle\frac{\mathrm{d}}{\mathrm{d}x}\tanh^{-1}(x)=\frac{1}{1-x^2}.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: separation of variables involving a quadratic
PostPosted: Wed, 12 Sep 2012 01:36:18 UTC 
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Hello, canadianicon25!

Quote:
\dfrac{dx}{dt} \:=\:k(3-x)(2-x)\quad x(0)\,=\,1

I have a worked solution for this problem and it uses partial fractions as a key method.

My question is whether this problem can be solved without partial fractions,
ie: combining terms to produce a quadratic in x.

A yes or no is what I am looking for.
If yes, I will attempt to work the problem.

We can also complete-the-square.

(x-3)(x-2) \;=\;x^2 - 5x + 6

. . . . . . . . . . . .=\:\left(x^2 - 5x + \frac{25}{4}\right) + \left(6 - \frac{25}{4}\right)

. . . . . . . . . . . .=\;\left(x - \frac{5}{2}\right)^2 - \left(\frac{1}{2}\right)^2


Then the left side has the form: .\displaystyle\int\frac{du}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right| + C



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 Post subject: Re: separation of variables involving a quadratic
PostPosted: Wed, 12 Sep 2012 01:57:25 UTC 
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Doesn't the derivation of that integration rule use partial fractions?


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 Post subject: Re: separation of variables involving a quadratic
PostPosted: Wed, 12 Sep 2012 06:38:29 UTC 
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Matt wrote:
Doesn't the derivation of that integration rule use partial fractions?


I think so, but it's probably in some table as well, which is probably what Soroban surpressed.

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