Solution of Separable Equation, Plotting Graph, Interval Estimation

[shamieh]

(a) Find the solution of the given initial value problem in explicit form.
(b) Plot the graph of the solution
(c) Determine (at least approximately) the interval in which the solution is defined

\(\displaystyle \frac{dr}{dx} =\frac{r^2}{x} \) and \(\displaystyle r(1) = 2\)

I'm kind of confused..How do I start this problem?

 

[MarkFL]

Well, the first thing you want to do is separate the variables:

\(\displaystyle \frac{1}{r^2}\,dr=\frac{1}{x}\,dx\)

Notice we just lost the trivial solution \(\displaystyle r\equiv0\), however we know this cannot be the solution to the IVP given the initial condition.

So, integrate both sides...what do you get?

 

[shamieh]

\(\displaystyle -\frac{1}{r} = ln(x) + C\)

now do I need to plug in r(1) = 2?

 

[MarkFL]

\(\displaystyle -\frac{1}{r} = ln(x) + C\)

now do I need to plug in r(1) = 2?

Yes, you could do that now, or you could solve for $r$ first...your choice. Another way to integrate is to use definite integrals along with your boundary conditions as the limits and change the dummy variables:

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

\(\displaystyle \left[-\frac{1}{u}\right]_2^r=\left[\ln|v|\right]_1^x\)

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

\(\displaystyle \frac{1}{r}=\frac{1}{2}-\ln|x|=\frac{1-\ln\left(x^2\right)}{2}\)

\(\displaystyle r=\frac{2}{1-\ln\left(x^2\right)}\)

 

[MarkFL]

I wanted to elaborate a little about the equivalence of using either indefinite integrals and then using the initial conditions to determine the constant of integration or using definite integrals and the boundaries as the limits of integration.

Suppose you have the initial value problem (IVP):

\(\displaystyle \frac{dy}{dx}=f(x)\) where \(\displaystyle y\left(x_0\right)=y_0\)

Now, separating variables and using indefinite integrals, we may write:

\(\displaystyle \int\,dy=\int f(x)\,dx\)

And upon integrating, we find

\(\displaystyle y(x)=F(x)+C\) where \(\displaystyle \frac{d}{dx}\left(F(x) \right)=f(x)\)

Using the initial condition, we get

\(\displaystyle y\left(x_0 \right)=F\left(x_0 \right)+C\)

Solving for $C$ and using \(\displaystyle y\left(x_0\right)=y_0\), we obtain:

\(\displaystyle C=y_0-F\left(x_0 \right)\) thus:

\(\displaystyle y(x)=F(x)+y_0-F\left(x_0 \right)\)

which we may rewrite as:

\(\displaystyle y(x)-y_0=F(x)-F\left(x_0 \right)\)

Now, we may rewrite this, using the anti-derivative form of the fundamental theorem of calculus, as:

\(\displaystyle \int_{y_0}^{y(x)}\,dy=\int_{x_0}^{x}f(x)\,dx\)

Now, since the variable of integration gets integrated out, it is therefore considered a "dummy variable" and since it is considered good form not to use the same variable in the boundaries as we use for integration, we may switch these dummy variables and write:

\(\displaystyle \int_{y_0}^{y(x)}\,du=\int_{x_0}^{x}f(v)\,dv\)

This demonstrates that the two methods are equivalent.

Using the boundaries (the initial and final values) in the limits of integration eliminates the need to solve for the constant of integration, and I find it a more intuitive and cleaner approach to separable initial value problems.

 

[shamieh]

\(\displaystyle -1/r = ln(x) + c\)

is going to turn into \(\displaystyle r = \frac{-1}{ln(x)+c}\) right?

 

[MarkFL]

\(\displaystyle -1/r = ln(x) + c\)

is going to turn into \(\displaystyle r = \frac{-1}{ln(x)+c}\) right?

Yes, that would be valid (although you want the absolute value of $x$ as the argument for your natural log function). Another approach would be to begin with:

\(\displaystyle -\frac{1}{r}=\ln|x|+C\)

Now multiply both sides by -1, and observe that an arbitrary constant negated is still just an arbitrary constant, so we may write:

\(\displaystyle \frac{1}{r}=C-\ln|x|\)

And so:

\(\displaystyle r=\frac{1}{C-\ln|x|}\)

How you choose to do it though is up to you, as long as you perform legal algebraic steps you will arrive at the correct answer. :D

 

[shamieh]

ahh I see...But I can;t plug in the Initial condition r(1) = 2 because if I do then ln(1) is not going to work right? so does x have to be > 1 ?

 

[MarkFL]

ahh I see...But I can;t plug in the Initial condition r(1) = 2 because if I do then ln(1) is not going to work right? so does x have to be > 1 ?

You can use $\ln(1)$ since:

\(\displaystyle \ln(1)=0\)

This comes from the fact that:

\(\displaystyle e^0=1\)