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Consider the differential equation $$x''+\frac{a}{t}x'+\frac{b}{t^2}x=0$$ for $t>0$ where $a,b\in\mathbb{R}$ are constants.

I need to show that $x:\mathbb{R^+}\to\mathbb{R}$ is a solution of this if and only if $y:\mathbb{R^+}\to\mathbb{R}$ defined by $y(\tau)=x(exp(\tau))$ is a solution of

$$y'' + (a - 1)y' + by = 0$$

I also need to know how to determine the general real solution of the first equation, distinguishing between different cases depending on the sign of $\omega = (a - 1)^2 - 4b$.

Any help with the first part of second part would be greatly appreciated.

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1 Answer

The second equation is derived from the first using the chain rule:

$$\frac{d x}{d t} = \frac{d x}{d \tau} \frac{d \tau}{d t}$$

Note that

$$\frac{d \tau}{d t} = \frac{1}{(dt/d\tau)} = e^{-\tau}$$

Then use

$$\frac{d^2 x}{d t^2} = \frac{d}{d \tau} \left (\frac{d x}{d \tau} \frac{d \tau}{d t} \right ) \frac{d \tau}{d t}$$

Substituting $y(\tau) = x(e^{\tau})$ and $t=e^{\tau}$, we get the stated equation:

$$y'' + (a-1) y' + b y = 0$$

where the prime on $y$ is taken to be differentiation with respect to $\tau$. To solve this equation, assume $y(\tau) = A e^{r \tau}$ and solve for $r$:

$$r^2 + (a-1)r + b = 0 \implies r_{\pm} = \frac{-(a-1) \pm \sqrt{(a-1)^2 - 4 b}}{2}$$

When $(a-1)^2 - 4 b>0$, then the solutions are pure exponential; when $(a-1)^2 - 4 b<0$, there is an oscillatory (i.e., sine and cosine) component. That is, in the former case

$$y(\tau) = e^{-(a-1) \tau/2} [A e^{(\sqrt{(a-1)^2 - 4 b}) \tau/2} + B e^{-(\sqrt{(a-1)^2 - 4 b}) \tau/2}]$$

and in the latter case

$$y(\tau) = e^{-(a-1) \tau/2} [A \cos{[(\sqrt{(a-1)^2 - 4 b}) \tau/2]} + B \sin{[(\sqrt{(a-1)^2 - 4 b}) \tau/2]}]$$

You can then get $x(t) = y(\log{t})$ and verify that it satisfies the original differential equation. That is, write

$$x(t) = A t^{r_+} + B t^{r_-}$$

Plugging this into the original equation, get that each of $r_+$ and $r_-$ must satisfy

$$r (r-1) + a r + b=0 \implies r^2 + (a-1) r + b=0$$

which I hope looks familiar.

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