The solution of (1+y^2) dx - xy dy = 0,y(1)=0 | vedclass

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(D) Given the differential equation $(1+y^2) dx - xy dy = 0$ with the initial condition $y(1)=0$. Rearranging the terms,we get $(1+y^2) dx = xy dy$. S

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$\sqrt{2}$

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(D) Given the differential equation $(1+y^2) dx - xy dy = 0$ with the initial condition $y(1)=0$. Rearranging the terms,we get $(1+y^2) dx = xy dy$. Separating the variables,we have $\frac{dx}{x} = \frac{y dy}{1+y^2}$. Integrating both sides,$\int \frac{dx}{x} = \int \frac{y dy}{1+y^2}$. Let $t = 1+y^2$,then $dt = 2y dy$,so $y dy = \frac{1}{2} dt$. Thus,$\ln|x| = \frac{1}{2} \ln|1+y^2| + C_1 = \ln|\sqrt{1+y^2}| + C_1$. This implies $x = c\sqrt{1+y^2}$. Using the condition $y(1)=0$,we substitute $x=1$ and $y=0$: $1 = c\sqrt{1+0^2} \implies c=1$. So,$x = \sqrt{1+y^2}$,which squares to $x^2 = 1+y^2$ or $x^2 - y^2 = 1$. This is the equation of a rectangular hyperbola where $a^2=1$ and $b^2=1$. The eccentricity $e$ of a hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{1}} = \sqrt{2}$.

The solution of $(1+y^2) dx - xy dy = 0$,$y(1)=0$ represents a conic. Its eccentricity is

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