Solve the differential equation.

(1 - x^2)v''(x) + 2(1 + x^2)v'(x) = 0

1 answer

To solve this differential equation, we will first rewrite it in standard form by dividing by (1 - x^2):

v''(x) + 2(1 + x^2)/(1 - x^2) * v'(x) = 0

Simplify the expression:

v''(x) + 2/(1 - x^2) * v'(x) = 0
v''(x) + 2/(1 - x^2) * v'(x) = 0

Now we will make the substitution u = v'(x) to rewrite the equation in terms of u:

u' + 2/(1 - x^2) * u = 0
u' = -2u/(1 - x^2)

Separate variables and integrate:

∫du/u = -2∫dx/(1 - x^2)
ln|u| = -2 arctan(x) + C
u = e^(-2 arctan(x) + C)
u = Ce^(-2 arctan(x))

Now substitute back v'(x) = Ce^(-2 arctan(x)):

v'(x) = Ce^(-2 arctan(x))

Integrate v'(x) to find v(x):

v(x) = ∫Ce^(-2 arctan(x)) dx

This integral may not have a simple closed-form solution, so the general solution to the differential equation is:

v(x) = Ce^(-2 arctan(x)) + D

where C and D are constants.