luissoccer4444pc2og8 luissoccer4444pc2og8
  • 18-07-2018
  • Mathematics
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solve the differential equations dy/dx=((xy^2)+x)/y

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konrad509
konrad509 konrad509
  • 18-07-2018

[tex] \displaystyle
\dfrac{dy}{dx}=\dfrac{xy^2+x}{y}\\
\dfrac{dy}{dx}=\dfrac{x(y^2+1)}{y}\\
\dfrac{y}{y^2+1}\, dy=x\, dx\\
\int \dfrac{y}{y^2+1}\, dy=\int x\, dx\\
\dfrac{\ln (y^2+1)}{2}=\dfrac{x^2}{2}+C\\
\ln(y^2+1)=x^2+C\\
y^2+1=e^{x^2+C}\\
y^2=e^{x^2+C}-1\\
y=\sqrt{e^{x^2+C}-1} \vee y=-\sqrt{e^{x^2+C}-1}\\
\boxed{y=\sqrt{Ce^{x^2}-1} \vee y=-\sqrt{Ce^{x^2}-1}}


[/tex]

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