Question

one of the solutiin of the equation (1-x^2) y''-2xy'+12y=0 is
Ans-P3(x)

Answer 1

Let $y=ax^3+bx^2+cx+d.$ Then

$$y'=3ax^2+2bx+c$$

$$y''=6ax+2b$$

Substitute

$$(1-x^2) (6ax+2b)-2x(3ax^2+2bx+c)$$

$$+12(ax^3+bx^2+cx+d)=0$$

$$(-6a-6a+12a)x^3+(-2b-4b+12b)x^2$$

$$+(6a-2c+12c)x+2b+12d=0$$

$x^3:0=0$

$x^2:6b=0$

$x^1: 6a+10c=0$

$x^0: 2b+12d=0$

$$c=-\dfrac{3}{5}a, b=0, d=0$$

$$y=ax^3-\dfrac{3}{5}ax, a\in\R$$

One of the solutiin of the given equation is

$$y=5x^3-3x$$