To solve the equation of X, x square + m times x square + 3x m is not equal to 1

To solve the equation of X, x square + m times x square + 3x m is not equal to 1

According to the solving process, 1 + m ≠ 0, m ≠ - 1
(1+m)X²+3X=0
X[(1+m)X+3]=0
X = 0 or (1 + m) x + 3 = 0
X1=0,X2=-3/(1+m).

Y = (2x + 2) / (x ^ 2 + 2x-3), find its n-order derivative

Given y = (x ^ 2-1) ^ n, it is proved that (n + 2) order derivative of (x ^ 2-1) * y + (n + 1) order derivative of 2x * y - n (n + 1) order derivative of n (n + 1) y = 0
Given y = (x ^ 2-1) ^ n, it is proved that the (n + 2) order derivative of (x ^ 2-1) * y + the (n + 1) order derivative of 2x * y - the n order derivative of n (n + 1) * y = 0
There is a multiply sign missing on it. I hope you can understand it.

From y = (x ^ 2-1) ^ n to y '= n (x ^ 2-1) ^ (n-1) * 2x = 2nx * (x ^ 2-1) ^ n / (x ^ 2-1) = 2nxy / (x ^ 2-1), we can get the following results: (x ^ 2-1) y' = 2nxy. In order to avoid confusion, if necessary, let the derivative order in double brackets to calculate the n-order derivatives at both ends respectively. By using Leibniz formula, we can get: ∑ [i = 0, n] C (n, I) (x ^