I'm quite confused at this question. There seems to be lots of different formulae floating around, but none of them say how to find the Wronskian with only one solution to the equation.
How to find the Wronskian when given one solution (reduction of order)
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ordinary-differential-equations
1 Answers
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Let $y_1=x$ and $y_2$ be the two solutions of the equation $$y''= \frac{2x}{1-x^2}y'-\frac{2}{1-x^2}y$$ The Wronskian formula is $$W(x)=y_1y_2'-y_2y_1'$$
Additionally $$W'(x)=\frac{2x}{1-x^2} \cdot W(x)$$
This is an ODE, which can be solved by seperation and we get: $$W(x)=e^{\int \frac{2x}{1-x^2} dx}=e^{-ln(1-x^2)}=-\frac{1}{x^2-1}$$
Now we know that $y_1=x$ is a solution and put it in the formula and form to: $$\frac{1}{x^2-1}\cdot\frac{1}{x}=y_2'-\frac{1}{x}y_2$$
Solve this to get: $$y_2=x+\frac{1}{2}\cdot x\cdot ln(1-x)-\frac{1}{2}\cdot x\cdot ln(1+x)+1$$
By the way:
For showing that $y_1=x$ is a solution, you have to plug $y_1$ in the euqation: $$y''= \frac{2x}{1-x^2}y'-\frac{2}{1-x^2}y$$ $y_1=x \Rightarrow y_1'=1 \Rightarrow y_1''=0$ $$0= \frac{2x}{1-x^2}\cdot 1-\frac{2}{1-x^2}\cdot x = 0$$