[sage-support] Re: C.I.L. by Integration problem
I'm told it will be in 2.8.3, whose release is planned
on Thursday or Friday.
+++
On 8/22/07, Ted Kosan <[EMAIL PROTECTED]> wrote:
>
> David wrote:
>
> > You can try this:
> >
> > sage: t = var('t')
> > sage: y = function('y', t)
> > sage: de = lambda y: diff(y,t) - (4/100)*y
> > sage: desolve_laplace(de(y(t)),["t","y"],[0,100])
> > '100*%e^(t/25)'
> >
> > This was just implemented.
>
> What version of Sage do you think desolve_laplace will be in? I just
> looked for this function in the public Notebook ( 'SAGE Version
> 2.8.2, Release Date: 2007-08-22' ) but it was not present.
>
> Thanks :-)
>
> Ted
>
> >
>
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[sage-support] Re: C.I.L. by Integration problem
David wrote:
> You can try this:
>
> sage: t = var('t')
> sage: y = function('y', t)
> sage: de = lambda y: diff(y,t) - (4/100)*y
> sage: desolve_laplace(de(y(t)),["t","y"],[0,100])
> '100*%e^(t/25)'
>
> This was just implemented.
What version of Sage do you think desolve_laplace will be in? I just
looked for this function in the public Notebook ( 'SAGE Version
2.8.2, Release Date: 2007-08-22' ) but it was not present.
Thanks :-)
Ted
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[sage-support] Re: C.I.L. by Integration problem
You can try this:
sage: t = var('t')
sage: y = function('y', t)
sage: de = lambda y: diff(y,t) - (4/100)*y
sage: desolve_laplace(de(y(t)),["t","y"],[0,100])
'100*%e^(t/25)'
This was just implemented.
+
On 8/18/07, Ted Kosan <[EMAIL PROTECTED]> wrote:
>
> Does anyone have any thoughts on how to solve the following problem in Sage?"
>
> A quantity y increases with x at a rate constantly equal to .04 y.
> If y = 100 at x = 0, find the formula.
>
>
> Here is the problem solved manually:
>
> """
> A quantity y increases with x at a rate constantly equal to .04 y.
> If y = 100 at x = 0, find the formula.
>
> Given:
> dy/dx = .04 * y
>
> The problem is to integrate this and obtain y in terms of x.
>
> A difficulty is that the derivative on the left side is taken with
> respect to x and the right member is expressed in terms of y.
>
> Divide through by y:
> 1/y * dy/dx = .04
>
> Express in differential notation.
> dy/y = .04 * dx
>
> Therefore:
> integral(dy/y) = integral(.04 * dx) + C
>
> And:
> log(y) = .04 * x + C
>
>
> y = 100 when x = 0, therefore C = log(100):
> log(y) = .04 * x + log(100)
>
> log(y) - log(100) = .04 * x
>
> log(y/100) = .04 * x
>
> This means that .04 * x is the exponent of the power to which the
> base e must be raised to equal the fraction y/100.
>
> Therefore:
> y/100 = e^(.04 * x)
>
> or:
> y = 100 * e^(.04 * x)
> """
>
> Thanks in advance :-)
>
> Ted
>
> >
>
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