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

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

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