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[tex]y' = \frac{1+2x+2y}{1-2x-2y}[/tex]

Hege Baggethun2020 skrev:[tex]y' = \frac{1+2x+2y}{1-2x-2y}[/tex]

[tex]y' = \frac{1+2x+2y}{1-2x-2y}=y' = \frac{1+2(x+y)}{1-2(x+y)}\\ \\der\,\,v=x+y\,=>\,\frac{dv}{dx}=1+\frac{dy}{dx}\\ \\ \\y'=\frac{dv}{dx}-1=\frac{1+2v}{1-2v}\\ \\\frac{dv}{dx}=\frac{1+2v+1-2v}{1-2v}=\frac{2}{1-2v}\\ \\ \\\int (1-2v)dv=2\int dx \\v-v^2=2x+c\\ (x+y)^2-(x+y)+(2x+c)=0\\ \\ (x+y)=\frac{1\pm\sqrt{1-4(2x+c)}}{2}\\ y(x)=y=-x+\frac{1\pm\sqrt{1-4(2x+c)}}{2}\\[/tex]

Janhaa skrev:

Hege Baggethun2020 skrev:[tex]y' = \frac{1+2x+2y}{1-2x-2y}[/tex]

[tex]y' = \frac{1+2x+2y}{1-2x-2y}=y' = \frac{1+2(x+y)}{1-2(x+y)}\\ \\der\,\,v=x+y\,=>\,\frac{dv}{dx}=1+\frac{dy}{dx}\\ \\ \\y'=\frac{dv}{dx}-1=\frac{1+2v}{1-2v}\\ \\\frac{dv}{dx}=\frac{1+2v+1-2v}{1-2v}=\frac{2}{1-2v}\\ \\ \\\int (1-2v)dv=2\int dx \\v-v^2=2x+c\\ (x+y)^2-(x+y)+(2x+c)=0\\ \\ (x+y)=\frac{1\pm\sqrt{1-4(2x+c)}}{2}\\ y(x)=y=-x+\frac{1\pm\sqrt{1-4(2x+c)}}{2}\\[/tex]

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