If the solution of the differential equation | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{1+x}{2y}$.Separating the variables,we get: $2y \, dy = (1+x) \, dx$.Integrating both sides

Vedclass · Accepted Answer

$\sqrt{\frac{3}{2}}$

Vedclass · Answer

(B) Given the differential equation: $\frac{dy}{dx} = \frac{1+x}{2y}$.Separating the variables,we get: $2y \, dy = (1+x) \, dx$.Integrating both sides: $\int 2y \, dy = \int (1+x) \, dx$.$y^2 = x + \frac{x^2}{2} + C$.Since the conic passes through $(1, 1)$,substitute $x=1$ and $y=1$: $1^2 = 1 + \frac{1^2}{2} + C \Rightarrow 1 = 1.5 + C \Rightarrow C = -0.5$.So,$y^2 = x + \frac{x^2}{2} - 0.5$.Rearranging: $\frac{x^2}{2} + x - y^2 = 0.5$.Multiply by $2$: $x^2 + 2x - 2y^2 = 1$.Complete the square for $x$: $(x^2 + 2x + 1) - 2y^2 = 1 + 1$.$(x+1)^2 - 2y^2 = 2$.Divide by $2$: $\frac{(x+1)^2}{2} - \frac{y^2}{1} = 1$.This is a hyperbola of the form $\frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1$ where $a^2 = 2$ and $b^2 = 1$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}$.

If the solution of the differential equation $\frac{dy}{dx} = \frac{1+x}{2y}$ is a conic passing through the point $(1, 1)$,then its eccentricity is:

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