решить дифференциальное уравнение(1+x^2)y''+(y')^2+1=0 y(0)=y'(0)=1

$(1+x^2)y''+(y')^2+1=0\\ y' = f(x)\\ (1+x^2)f'+f^2+1=0\\ (1+x^2)\frac{df}{dx}+f^2+1=0\\ \frac{df}{1+f^2} = -\frac{dx}{1+x^2}\\ \int \frac{df}{1+f^2} = -\int \frac{dx}{1+x^2} + C_1\\ \arctan f = -\arctan x + C_1\\ f = \tan(-\arctan x + C_1) = \frac{\tan C_1 - \tan (\arctan x)}{1+\tan C_1\tan(\arctan x)}\\ y' = \frac{C_2-x}{1+xC_2}$

$y'(0) = 1 = C_2\\ y' = \frac{1-x}{1+x}\\ y = \int \frac{1-x}{1+x}dx = -\int \frac{x-1}{x+1}dx=-\int (1 - \frac{2}{x+1})dx=\\ =-x+\int \frac{2dx}{x+1} = 2\ln|x+1| - x + C_3\\ y(0) = 1 = C_3\\ y = 2\ln|x+1| - x + 1$