The slope of the tangent at any point (x, y) | vedclass

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(B) The differential equation is given by $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$.Dividing numerator and denominator by $x^2$,we get $\frac{dy}{dx} = \f

Vedclass · Accepted Answer

$4 \sqrt{3}$

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(B) The differential equation is given by $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$.Dividing numerator and denominator by $x^2$,we get $\frac{dy}{dx} = \frac{1+(y/x)^2}{2(y/x)}$.Let $y = tx$,then $\frac{dy}{dx} = t + x\frac{dt}{dx}$.Substituting this into the equation: $t + x\frac{dt}{dx} = \frac{1+t^2}{2t}$.$x\frac{dt}{dx} = \frac{1+t^2}{2t} - t = \frac{1-t^2}{2t}$.Separating variables: $\int \frac{2t}{1-t^2} dt = \int \frac{dx}{x}$.Integrating both sides: $-\ln|1-t^2| = \ln|x| + C$,which implies $\ln|1-t^2|^{-1} = \ln|cx|$.So,$\frac{1}{1-t^2} = cx$,or $1-t^2 = \frac{1}{cx}$.Substituting $t = y/x$: $1 - \frac{y^2}{x^2} = \frac{1}{cx} \Rightarrow \frac{x^2-y^2}{x^2} = \frac{1}{cx} \Rightarrow x^2-y^2 = \frac{x}{c}$.Given $y(2) = 0$,we have $2^2 - 0^2 = \frac{2}{c} \Rightarrow 4 = \frac{2}{c} \Rightarrow c = \frac{1}{2}$.The equation becomes $x^2 - y^2 = 2x$.At $x = 8$: $8^2 - y^2 = 2(8) \Rightarrow 64 - y^2 = 16 \Rightarrow y^2 = 48$.Thus,$y = \pm \sqrt{48} = \pm 4\sqrt{3}$.

The slope of the tangent at any point $(x, y)$ on a curve $y = y(x)$ is $\frac{x^2+y^2}{2xy}$,where $x > 0$. If $y(2) = 0$,then a value of $y(8)$ is

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