Question

1) y'=$\frac{x+1}{y-1}$

2) (1+$e^{x}$)y*y'=$e^{x}$

3) (x-3)dy +(2y-1)dx=0

Answer 1

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

$3)\; \; (x-3)dy+(2y-1)dx=0\; |:(x-3)(2y-1)\ne0\\\\\frac{dy}{2y-1}+\frac{dx}{x-3}=0\; ,\; \; \; \int \frac{dy}{2y-1}=-\int \frac{dx}{x-3}\\\\ \frac{1}{2}\cdot ln|2y-1|=-ln|x-3|+ln|C|\\\\ln\sqrt{|2y-1|}+ln|x-3|=ln|C|\\\\ln\sqrt{|2y-1|}\cdot |x-3|=ln|C|\\\\\sqrt{|2y-1|}\cdot |x-3|=C_1\; ,\; \; |C|=C_1\\\\\sqrt{|2y-1|}= \frac{C_1}{|x-3|} \\\\|2y-1|=\frac{C_1^2}{(x-3)^2}\; ,\; \; \; C_1^2=C_2\\\\|2y-1|= \frac{C_2}{(x-3)^2}$