Найти общее решение или общий интеграл данных дифференциальных уравнений первого порядка 1) y'+xy=xy^2 2) y^2-4xy+4x^2'=0 3)x (x-1)y'+2xy=1

[latex]1) y'+xy=xy^2[/latex] [latex]y'=xy^2-xy[/latex] [latex]y'=x(y-1)*y[/latex] [latex]\frac{y'}{(y-1)*y}=x[/latex] [latex]\int{\frac{y'}{(y-1)*y}}\,dx=\int{x}\,dx[/latex] [latex]ln|-y+1|-ln|y|=\frac{x^2}{2}+C_1[/latex] [latex]y=\frac{1}{e^{\frac{x^2}{2}+C_1}+1}[/latex] [latex]y=\frac{1}{C_1e^{\frac{x^2}{2}}+1}[/latex] [latex]2)y^2-4xy+4x^2y'=0[/latex] [latex]4x^2y'-4xy=-y^2[/latex] [latex]-\frac{y'}{y^2}+\frac{1}{xy}=\frac{1}{4x^2} [/latex] [latex]v=\frac{1}{y},togda\hspace*{10}v'=-\frac{y'}{y^2}[/latex] [latex]v'+\frac{v}{x}=\frac{1}{4x^2}[/latex] [latex]\mu=e^{\int{\frac{1}{x}}\,dx}=x[/latex] [latex]xv'+v=\frac{1}{4x}[/latex] [latex]1=x':[/latex] [latex]xv'+x'v=\frac{1}{4x}[/latex] [latex](xv)'=\frac{1}{4x}[/latex] [latex]\int{(xv)'}\,dx=\int{\frac{1}{4x}}\,dx[/latex] [latex]xv=\frac{ln|x|}{4}+C_1[/latex] [latex]v=\frac{\frac{ln|x|}{4}+C_1}{x}[/latex] [latex]y=\frac{1}{v}=\frac{4x}{ln|x|+4C_1}[/latex] [latex]y=\frac{4x}{ln|x|+C_1}[/latex] [latex]x(x-1)y'+2xy=1[/latex] [latex]y'+\frac{2y}{x-1}=\frac{1}{x(x-1)}[/latex] [latex]\mu=e^{\int{\frac{2}{x-1}}\,dx}=(x-1)^2[/latex] [latex](x-1)^2y'+2(x-1)y=\frac{x-1}{x}[/latex] [latex]2(x-1)=((x-1)^2)':[/latex] [latex](x-1)^2y'+((x-1)^2)y=\frac{x-1}{x}[/latex] [latex]((x-1)^2y)'=\frac{x-1}{x}[/latex] [latex]\int{((x-1)^2y)'}\,dx=\int{\frac{x-1}{x}}\,dx[/latex] [latex](x-1)^2y=x-ln|x|+C_1[/latex] [latex]y=\frac{x-ln|x|+C_1}{(x-1)^2}[/latex]