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Answer 1

Ответ:

$a)\ \ \displaystyle xy'=y\ \ ,\ \ \frac{dy}{dx}=\frac{y}{x}\ \ ,\ \ \ \int \frac{dy}{y}=\int \frac{dx}{x}\\\\ln|y|=ln|x|+lnC\ \ ,\ \ \ \ \underline {\ y=Cx\ }\\\\\\b)\ \ 2(xy+y)\, y'+x(y^4+1)=0\ \ ,\ \ \ y'=-\frac{x(y^4+1)}{ 2y(x+1)}\ \ ,\\\\\int 2y\, dy=-\int \frac{x\, dx}{x+1}\ \ ,\ \ \ y^2=-\int \Big(1-\frac{1}{x+1}\Big)\, dx\ \ ,\\\\y^2=-(x-ln|x+1|)+C \ \ ,\ \ \ \ \underline {\ y^2=ln|x+1|-x+C\ }$

$c)\ \ y-xy'=3(1+x^2y')\ \ ,\ \ \ y-xy'=3+3x^2y'\ \ ,\ \ y'(3x^2+x)=y-3\ ,\\\\\displaystyle \int \frac{dy}{y-3}=\int \frac{dx}{3x^2+x}\ \ ,\ \ \ ln|y-3|=\int \frac{dx}{x(3x+1)}\ ;\\\\\int \frac{dx}{x(3x+1)}=\int \Big(\ \frac{1}{x}-\frac{3}{3x+1}\, \Big)\, dx=ln|x|-ln|3x+1|+lnC_1\\\\ln|y-3|=ln|x(3x+1)|+lnC\\\\\underline{\ y-3=Cx(3x+1)\ }$