Finding the special solution of differential equation y '- 2Y / (1-x ^ 2) = x + 1, x = 0, y = 0

The integral factor is exp (∫ - 2 / (1-x ^ 2) DX) = (x-1) / (x + 1) two sides of differential equation multiply (x-1) / (x + 1) at the same time, then (x-1) / (x + 1) * y '+ 2 * y / (x + 1) ^ 2 = X-1, i.e. ((x-1) / (x + 1) * y' = X-1 two sides of differential equation are integrated, and (x-1) / (x + 1) * y = 1 / 2 * x ^ 2-x, then y = 1 / 2 * x * (X-2) * (x + 1) / (x-1)

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