Дифференциальные уравнения. За них даю 60б.

Решите задачу:

$1)\; \; (x^2+x)y'=2y+1\; \; ,\; \; \; y'=\frac{dy}{dx}\\\\ \frac{dy}{dx} = \frac{2y+1}{x^2+x} \\\\ \int \frac{dy}{2y+1}= \int \frac{dx}{x(x+1)}\\\\\frac{1}{x(x+1)}= \frac{A}{x}+ \frac{B}{x+1} \; \; \to \; \; 1=A(x+1)+Bx\\\\1=(A+B)x+A\; \; \Rightarrow \\\\x^0|\; 1=A\\\\x|\; 0=A+B\; \; \to \; \; \; B=-A=-1\\\\\int \frac{dx}{x(x+1)}=\int \frac{dx}{x} -\int \frac{dx}{x+1}=ln|x|-ln|x+1|+ln|C_1| \\\\\int \frac{dy}{2y+1}=\frac{1}{2}\int \frac{2\, dy}{2y+1} = \frac{1}{2}\int \frac{d(2y+1)}{2y+1}=\frac{1}{2}\, ln|2y+1|+ln|C_2|$

$\frac{1}{2}\, ln|2y+1|=ln|x|-ln|x+1|+ln|C|\\\\ln\sqrt{2y+1}=ln\frac{Cx}{x+1}\\\\\sqrt{2y+1}=\frac{Cx}{x+1}\\\\2)\; \; (1+x^2)y'+y(1+x^2)=xy\\\\(1+x^2)y'=y(x-1-x^2)\\\\ \frac{dy}{dx}=\frac{-y(x^2-x+1)}{1+x^2}\; \; \; [\; d(x^2+1)=(x^2+1)'dx=2x\, dx\; ]\\\\\int \frac{dy}{y}=-\int \frac{(x^2+1)-x}{x^2+1}dx\\\\\int \frac{dy}{y}=-\int (1-\frac{x}{x^2+1})dx\\\\ln|y|=-x+\frac{1}{2}\int \frac{d(x^2+1)}{x^2+1}\\\\ln|y|=-x+\frac{1}{2}\, ln|x^2+1|+C$