Linearly independent solutions of a differential?

Question

Considering the differential equation $2x^2y''+3xy' - y = 0.$

Determine $r\in\textbf{R}_{>0}$, such that $ y_2(x)=x^r$ be linearly independent of $y_1(x)= 1/x $.

I know that I must calculate $W(x)$ so that it is different from zero but I can't find the possible values ​​of $r$.

Can you help me please? Thanks!!

Answer 1

$$2x^2y''+3xy' - y = 0.$$ $$y''+\frac 3 {2x}y' -\frac 1 {2x^2} y = 0.$$ $$y''+ \color {blue} {p(x)}y'+q(x) y = 0.$$ The wronskian is easy to calculate: $$W=c \exp ( {-\int \color {blue} {p(x)} dx})$$ $$W=c \exp ( {-\int \frac 3 {2x}dx})$$ $$W=c x^{-3/2}$$ Since we already have a solution $y_1=\frac 1 x$ and $y_2=x^r$ : $$y_1y'_2-y'_1y_2=W$$ $$\frac 1 x rx^{r-1}+\frac 1 {x^2}x^r=c x^{-3/2}$$ $$ rx^{r}+x^r=c x^{1/2}$$ $$ (r+1)x^{r}=cx^{1/2}$$ $$ \implies r=\frac 12 \implies y_2(x)= x^{\frac 12 }$$ Since the differential equation is a Cauchy-Euler's equation it's easier to solve it with $y=x^r$ directly without the Wronskian.

Answer 2

Since it is a Cauchy-Euler ODE, we can solve it like this:

$$y(x) = x^r$$ Substitute in the ODE and you will get:

$$x^r(2r^2+r-1)=0$$

Let's just solve the quadratic and we then have:

$$r=-1,\frac{1}{2}$$

Therefore, your solution is $r=\frac{1}{2}$.