# Solving ODE $\alpha_0''(r) +1/r\space \alpha_0'(r) = -\frac{4\sqrt{r}}{3}\pi^2$

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.

• @Amzoti - Thanks for the feedback, this might be obvious, but where does the $\mathrm{ln}( r)$ term come from? – PolandAspect Oct 12 '14 at 16:46
• @Amzoti - Oh right, thanks, so if I was prescribed initial conditions I would use definite integrals in the integrating factors? – PolandAspect Oct 12 '14 at 16:52
• @Amzoti - So is there any context in which I would need to use definite integrals in the integrating factors? – PolandAspect Oct 12 '14 at 16:57
• These are also worth going through: howellkb.uah.edu/DEtext/Part2/linear.pdf (see Section 5.3). – Amzoti Oct 12 '14 at 17:09
• 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 – Alan Oct 12 '14 at 18:09

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 $I$, $$\alpha(r) = -\tfrac{16}{75}\pi^2r^2\sqrt{r}+a\log(r)+b.$$