MHB -2.2.12 dr/dθ = r^2/θ, r(1) = 2 IVP, graph, interval

[karush]

$$\d{r}{\theta}=\frac{r^2}{\theta},\quad r(1)=2$$
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

 

[MarkFL]

$$\d{r}{\theta}=\frac{r^2}{\theta},\quad r(1)=2$$
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:

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

What's the next step?

 

[karush]

I would begin with:<br>
<br>
$$\int_2^r u^{-2}\,du=\int_1^{\theta} v^{-1}\,dv$$<br>
<br>
What's the next step?

<br>
<br>
$$\frac{1}{u}\biggr|_2^r=\ln u\biggr|_1^\theta$$<br>
<br>
sorta maybe

 

[MarkFL]

$$\frac{1}{u}\biggr|_2^r=\ln u\biggr|_1^\theta$$

sorta maybe

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

 

[karush]

ok appreciate
ill be in Hamilton library tmro
to do more
just have a tablet at home which is very hard to use

 

[karush]

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 textbook 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

 

[MarkFL]

\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 textbook 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:

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

If we combine terms on the RHS:

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

Invert both sides:

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

Now we know for the log function:

$$0<\theta$$

And we know:

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

$$1\ne2\ln(\theta)$$

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

$$\theta\ne\sqrt{e}$$

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

$$0<\theta<\sqrt{e}$$

 

[karush]

Mahalo

I would have never gotten the interval

 

[karush]

https://www.physicsforums.com/attachments/8666
here is what I will turn in
quess some got lost in the transparency transform