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Artig differensiallikning
Posted: 18/06-2020 10:31
by Hege Baggethun2020
[tex]y' = \frac{1+2x+2y}{1-2x-2y}[/tex]
Re: Artig differensiallikning
Posted: 18/06-2020 11:37
by Janhaa
Hege Baggethun2020 wrote:[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]
Re: Artig differensiallikning
Posted: 31/07-2020 15:57
by Hege Baggethun2020
Janhaa wrote:Hege Baggethun2020 wrote:[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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