using Liouville-Ostrogradsky to solve differential equation

Question

I have a differential equation: $$y''+p_1\left(x\right)y'+p_0\left(x\right)y=0$$ for $x>0$ and it is given that $p_0,p_1$ are continuous for any $x\in \mathbb{R}$.

I am questioned if $y_1(x)=x$ and $y_2(x)=\ln(x)$ can be the solutions of this equation.

Answer: They are linear independent, therefore using Liouville-Ostrogradsky: $$W(y_1,y_2)(x)=ce^{-\int \:p_1\left(x\right)dx}$$ we get: $$W(y_1,y_2)(x)=\left|\begin{pmatrix}x&\ln\left(x\right)\\ 1&\frac{1}{x}\end{pmatrix}\right|=\ln\left(x\right)-1$$ so, we can choose c=1 for comfort and get: $$\ln\left(x\right)-1=e^{-\int p_1\left(x\right)dx}$$ therefore: $$\int p_1\left(x\right)dx=\frac{1}{\ln\left(\ln\left(x\right)-1\right)}$$ Therefore, after deriving both sides: $$p_1\left(x\right)=\frac{1}{x\cdot \left(\ln\left(x\right)-1\right)\cdot \ln^2\left(\ln\left(x\right)-1\right)}$$

Now the thing that bothers me, that is was said about $p_1$ that it is continuous for $x\in \mathbb{R}, x>0$, and here the denominator is defined only for any $x>e$, so I don't know if the authors of the question meant that it should be continuous for any $x>e$ and then we can continue and find $p_0$ or it contradicts the given statement and therefore those two functions cannot be a solution for the equation.

Am I right that those are the two options of answers for this diff. equation ? If it is about $x>e$, then we can choose $$y=x+\ln(x)$$, calculate $y'$ and $y''$, put it all together in the first formula: $y''+p_1\left(x\right)y'+p_0\left(x\right)y=0$ and extract $p_0$.

Otherwise if the question means that $p_1(x)$ has to be continuous on $x>0$, then $p_1(x)$ doesn't satisfies that because it is defined only for $x>e$, is this correct ?

Answer 1

If $t_1,y_2$ are two linearly independent solutions of a second order ODE. This ODE is given by: $$\begin{vmatrix} y'' & y'& y \\ y_1'' & y_1' & y_1 \\ y_2'' & y_2' & y_2 \end{vmatrix}=0$$ Here we have$$\begin{vmatrix} y'' & y'& y \\0 & 1 & x \\ -1/x^2 & 1/x & \ln x \end{vmatrix}=0$$ $$\implies (\ln x-1)y''-\frac{1}{x}y'+\frac{1}{x^2}y=0$$ So $$p_1(x)=(x(1-\ln x))^{-1}, p_0(x)=(x^2(\ln x-1))^{-1}$$

Answer 2

You could short-circuit the task from the start. Any solution, if not artificially restricted, has the whole of $\Bbb R$ as domain, as the coefficients are assumed as continuous everywhere. With the restriction to the positive half axis, this still gives a finite limit for $x\to0$, as in the extended solution this also is a point of continuity. Thus you can not get a solution $y=\ln x$ or $y=\ln|x|$ with its singularity at $x=0$.