MHB Change of Variables: Diff Eqn $y''+ \frac{p}{x} y'+ \frac{q}{x^2}y=0$

[evinda]

Hello! (Smile)

We have the differential equation $y''+ \frac{p}{x} y'+ \frac{q}{x^2}y=0, x>0$ and we set $z=\log x$.

Then $y'=\frac{dy}{dx}=\frac{dy}{dz} \frac{dz}{dx}=\frac{1}{x} \frac{dy}{dz}$

$y''=\frac{d^2y}{dx^2}=\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dx}\left( \frac{1}{x} \frac{dy}{dz} \right)=-\frac{1}{x^2}\frac{dy}{dx}+\frac{1}{x} \frac{d}{dx}\left( \frac{dy}{dz}\right)$

How can we find $\frac{d}{dx}\left( \frac{dy}{dz}\right)$?

 

[evinda]

We use the fact that $\frac{d}{dx} = \frac{dz}{dx}\frac{d}{dz}$.

Then we have the differential equation:

$$\frac{d^2y}{dz^2}+(p-1) \frac{dy}{dz}+qy=0 (*)$$

We are looking for a solution of the form $e^{rz}$.

$e^{rz}$ is a solution iff

$$r^2+(p-1)r+q=0$$

If $r_1, r_2$ two distinct real roots then $y_1(z)=e^{r_1 z}, y_2(z)=e^{r_2 z}$ are linearly indepent solutions of $(*)$ and so $y_1(x)=x^{r_1}, y_2(x)=x^{r_2}$ are linearly independent solutions of $y''+\frac{p}{x}y'+\frac{q}{x^2}y=0, x>0$.

How do we get that $y_1(x)=x^{r_1}, y_2(x)=x^{r_2}$?

 

[evinda]

It holds that $y_1(z)=e^{r_1 z}=(e^{z})^{r_1}=(e^{\log x})^{r_1}=x^{r_1}$.So $y_1(\log x)= x^{r_1}$.How do we get that $y_1(x)=x^{r_1}$ ?