Question

Find the Wronskian of a given set {y1, y2} of solutions of
(1 − x^2)y" − 2xy' + α(α + 1)y = 0 when W(0) = 1

Answer 1

Solution.

(1-x²)y'' - 2xy' + α(α + 1)y = 0

Dividing both sides of the equation by (1-x²) we get differential equation in standard form:

$y''-\frac{2x}{1-x^2}y'+\frac{\alpha \left(\alpha +1\right)}{1-x^2}y=0$ ;

p(t) is the coefficient of y' and q(t) is the coefficient of y:

$p\left(t\right)=-\frac{2x}{1-x^2}$

$q\left(t\right)=\frac{\alpha \left(\alpha +1\right)}{1-x^2}$

To determine the Wronskian we apply Abel's theorem:

$W\left(y_1,y_2\right)\left(x\right)=C\cdot e^{\left(-\int p\left(x\right)dx\right)}$

$W\left(y_1,y_2\right)=C\cdot e^{\left(-\int \left(-\frac{2x}{1-x^2}dx\right)\right)}$

$W\left(y_1,y_2\right)=C\cdot e^{\left(-\int \left(\frac{d\left(1-x^2\right)}{1-x^2}\right)\right)}$

$W\left(y_1,y_2\right)=C\cdot e^{-\ln \left|1-x^2\right|}$

$W\left(y_1,y_2\right)=\frac{C}{\left|1-x^2\right|}$

when W(0) = 1, we will get:

$1=\frac{C}{\left|1-0\right|}$

C = 1, thus:

$W\left(y_1,y_2\right)=\frac{1}{\left|1-x^2\right|}$

Answer: W = 1 / |1 - x²|