Differential Equations solution help

[Ted123]

One solution of the differential equation

$x^2(x^2+1)y^{\prime\prime} - 2x^3 y^{\prime} + 2(x^2-1) y = 0$

can be obtained in the form $y_1 = x^n$. Use this solution to find another, and in this way find the general solution.

The DE can be written as:

$\displaystyle y^{\prime\prime} - \frac{2x}{x^2+1} y^{\prime} + \frac{2(x^2-1)}{x^2(x^2+1)}y = 0$.


$\displaystyle - \int \frac{2x}{x^2+1} = -\ln (x^2+1)$

Therefore the Wronskian $\displaystyle W(x) = Ce^{\ln (x^2+1)} = C(x^2+1)$.

By inspection $y_1 = x^2$ - how can you see this straight away?

To find $y_2$ use the formula below with $W$ for $C=1$ - can you always just take $C=1$?

$\displaystyle y_2 = y_1 \int \frac{W}{y_1^2}\;dx$

 

[HallsofIvy]

Yes, you can but you could use other numbers as well- there are, after all, an infinite number of solutions.