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

• MHB
• karush
In summary: 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<\thetaAnd we know:1-2\ln(\theta)\ne01\ne2\ln(\theta)\frac{1}{2}\ne\ln(\
karush
Gold Member
MHB
$$\d{r}{\theta}=\frac{r^2}{\theta},\quad r(1)=2$$
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:
karush said:
$$\d{r}{\theta}=\frac{r^2}{\theta},\quad r(1)=2$$
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?

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

sorta maybe

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

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

MarkFL said:
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: karush said: \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: \(\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:

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

Mahalo

I would have never gotten the interval

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

## 1. What does the given equation represent?

The given equation is a first-order ordinary differential equation, with an initial value condition (IVP). It represents the rate of change of r with respect to θ, which is equal to r squared divided by θ. The equation also includes an initial condition, where r(1) = 2.

## 2. How do you solve this differential equation?

To solve this differential equation, you can use the method of separation of variables. This involves isolating the variables on each side of the equation and integrating both sides. You can then use the initial condition to find the specific solution.

## 3. What is the graph of this differential equation?

The graph of this differential equation will be a curve in the r-θ plane. The specific shape of the curve will depend on the initial condition and the interval of θ values. The curve will approach the r-axis as θ approaches infinity.

## 4. What is the interval of θ values for this differential equation?

The interval of θ values for this differential equation will depend on the initial condition. In this case, r(1) = 2, so the interval of θ values will be from 0 to infinity. However, if the initial condition was different, the interval of θ values could also be different.

## 5. Can you provide an example of a real-world application of this differential equation?

One example of a real-world application of this differential equation is in population growth modeling. The equation can represent the growth rate of a population (r) with respect to time (θ), with the initial condition representing the initial population size. This type of model can be used in various fields, such as biology, economics, and ecology.

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