0
$\begingroup$

How would one go about solving the ODE

$$\alpha''(r) +\frac{ \alpha'(r)}{r} = -\frac{4\sqrt{r}}{3}\pi^2$$

I know that one should use integrating factors, but I always get confused about them - when should the integrating factor involve a definite integral or when should it be indefinite?

I think that the integrating factor to be used here should be: $I(r) = exp(\int \frac{1}{r}dr) = r $

But I'm not sure if the integral should be definite or indefinite, then:

$$\frac{d}{dr}[\alpha'(r) \space r] = \frac{4r^{3/2}}{3}\pi^2$$

Integrating twice, I obtain $\alpha_0(r)= -\frac{16 \pi^2}{75} r^{5/2} $

Is this the most general solution, do I have to use definite integrals in my steps?

Sorry if this is really basic stuff, I can crank the handle on the algebra, but struggle with the theory of differential equations.

$\endgroup$
5
  • $\begingroup$ @Amzoti - Thanks for the feedback, this might be obvious, but where does the $ \mathrm{ln}( r) $ term come from? $\endgroup$ – PolandAspect Oct 12 '14 at 16:46
  • $\begingroup$ @Amzoti - Oh right, thanks, so if I was prescribed initial conditions I would use definite integrals in the integrating factors? $\endgroup$ – PolandAspect Oct 12 '14 at 16:52
  • $\begingroup$ @Amzoti - So is there any context in which I would need to use definite integrals in the integrating factors? $\endgroup$ – PolandAspect Oct 12 '14 at 16:57
  • $\begingroup$ These are also worth going through: howellkb.uah.edu/DEtext/Part2/linear.pdf (see Section 5.3). $\endgroup$ – Amzoti Oct 12 '14 at 17:09
  • $\begingroup$ Note that although this equation appears to be a second order ODE, it's actually a first order ODE in $\alpha '$, so treat $\alpha '$ as what you are solving for with first order techniques, then integrate that $\endgroup$ – Alan Oct 12 '14 at 18:09
0
$\begingroup$

Let us use the following simple but key result:

If $u'=v'$ on some interval $I$ then there exists some constant $c$ such that $u(x)=v(x)+c$ for every $x$ in $I$.

The differential equation to be solved is defined on $I=(0,+\infty)$ and you showed yourself that it is equivalent to $$[\alpha'(r) \space r]' = -\tfrac43\pi^2r\sqrt{r},$$ which can be rewritten as $$[\alpha'(r) \space r]' =\left[-\tfrac8{15}\pi^2r^2\sqrt{r}\right]'.$$ By a first application of the key result, there exists some constant $a$ such that $$\alpha'(r) \space r = -\tfrac8{15}\pi^2r^2\sqrt{r}+a,$$ that is, $$\alpha'(r) = -\tfrac8{15}\pi^2r\sqrt{r}+\frac{a}r=\left[-\tfrac{16}{75}\pi^2r^2\sqrt{r}+a\log r\right]'.$$ By a second application of the key result, there exists some constant $b$ such that, for every $r$ in $I$, $$\alpha(r) = -\tfrac{16}{75}\pi^2r^2\sqrt{r}+a\log(r)+b.$$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.