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

#### karush

##### Well-known member
$$\d{r}{\theta}=\frac{r^2}{\theta},\quad r(1)=2$$
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

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#### MarkFL

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

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#### MarkFL

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

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

#### karush

##### Well-known member
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
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??
\$(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 \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
Mahalo

I would have never gotten the interval

#### karush

##### Well-known member
View attachment 8666
here is what I will turn in
quess some got lost in the transparency transform