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[SOLVED] de2.2.12 dr/dθ = r^2/θ, r(1) = 2 IVP, graph, interval

karush

Well-known member
Jan 31, 2012
2,886
$$\d{r}{\theta}=\frac{r^2}{\theta},\quad r(1)=2$$
from i would deduct that $dr=r^2$ and $d\theta = \theta then$
$$\d{r}{\theta}=\frac{\theta}{r^2}
\text{ or }
\frac{1}{r^2}dr=\frac{1}{\theta}d\theta$$
intregrate
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$$\d{r}{\theta}=\frac{r^2}{\theta},\quad r(1)=2$$
from i would deduct that $dr=r^2$ and $d\theta = \theta!$ then
$$\d{r}{\theta}=\frac{\theta}{r^2}$$
but to intregrate ??
I would begin with:

\(\displaystyle \int_2^r u^{-2}\,du=\int_1^{\theta} v^{-1}\,dv\)

What's the next step?
 

karush

Well-known member
Jan 31, 2012
2,886
I would begin with:<br>
<br>
\(\displaystyle \int_2^r u^{-2}\,du=\int_1^{\theta} v^{-1}\,dv\)<br>
<br>
What's the next step?
<br>
<br>
$$\frac{1}{u}\biggr|_2^r=\ln u\biggr|_1^\theta$$<br>
<br>
sorta maybe
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$$\frac{1}{u}\biggr|_2^r=\ln u\biggr|_1^\theta$$

sorta maybe
You want a negative sign on the left resulting from the application of the power rule. :)
 

karush

Well-known member
Jan 31, 2012
2,886
ok appreciate
ill be in Hamilton library tmro
to do more
just have a tablet at home which is very hard to use
 

karush

Well-known member
Jan 31, 2012
2,886
You want a negative sign on the left resulting from the application of the power rule. :)
\begin{align*}\displaystyle
\frac{1}{u}\biggr|_2^r&= -\ln u\biggr|_1^\theta\\
\frac{1}{r}-\frac{1}{2}&=-\biggr[\ln{\theta}-\ln 1\biggr]=\ln \theta\\
\frac{1}{r}&=-\ln \theta+\frac{1}{2}
\end{align*}
well so far??
the text book answer is
$(a)\quad \displaystyle r = \frac{2}{1 − 2\, \ln θ} \\(c)\quad 0 < θ <
\sqrt{e}$

i continued but couldn't get this answer
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\begin{align*}\displaystyle
\frac{1}{u}\biggr|_2^r&= -\ln u\biggr|_1^\theta\\
\frac{1}{r}-\frac{1}{2}&=-\biggr[\ln{\theta}-\ln 1\biggr]=\ln \theta\\
\frac{1}{r}&=-\ln \theta+\frac{1}{2}
\end{align*}
well so far??
the text book answer is
$(a)\quad \displaystyle r = \frac{2}{1 − 2\, \ln θ} \\(c)\quad 0 < θ <
\sqrt{e}$

i continued but couldn't get this answer
You have:

\(\displaystyle \frac{1}{r}=-\ln(\theta)+\frac{1}{2}\)

If we combine terms on the RHS:

\(\displaystyle \frac{1}{r}=\frac{1-2\ln(\theta)}{2}\)

Invert both sides:

\(\displaystyle r=\frac{2}{1-2\ln(\theta)}\)

Now we know for the log function:

\(\displaystyle 0<\theta\)

And we know:

\(\displaystyle 1-2\ln(\theta)\ne0\)

\(\displaystyle 1\ne2\ln(\theta)\)

\(\displaystyle \frac{1}{2}\ne\ln(\theta)\)

\(\displaystyle \theta\ne\sqrt{e}\)

As \(1<\sqrt{e}\), and we need the part of the solution containing the initial value, we conclude:

\(\displaystyle 0<\theta<\sqrt{e}\)
 

karush

Well-known member
Jan 31, 2012
2,886
Mahalo

I would have never gotten the interval
 

karush

Well-known member
Jan 31, 2012
2,886
View attachment 8666
here is what I will turn in
quess some got lost in the transparency transform