The slope of the tangent at (x, y) to the cur | vedclass

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Question

(A) The slope of the tangent is given by $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$.
This is a homogeneous differential equation.
Let $y = vx$,then $\fra

Vedclass · Accepted Answer

$2y^2 = 2x^2 - 3x$

Vedclass · Answer

(A) The slope of the tangent is given by $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$.
This is a homogeneous differential equation.
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x^2 + v^2x^2}{2x(vx)} = \frac{1+v^2}{2v}$.
$x \frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1+v^2-2v^2}{2v} = \frac{1-v^2}{2v}$.
Separating variables: $\frac{2v}{1-v^2} dv = \frac{dx}{x}$.
Integrating both sides: $-\ln|1-v^2| = \ln|x| + C$.
$\ln|1-v^2|^{-1} = \ln|x| + C \implies \frac{1}{1-v^2} = Cx$.
Substituting $v = \frac{y}{x}$: $\frac{1}{1-(y/x)^2} = Cx \implies \frac{x^2}{x^2-y^2} = Cx \implies x^2 = C x (x^2-y^2) \implies x = C(x^2-y^2)$.
Since the curve passes through $(2, 1)$,$2 = C(4-1) = 3C \implies C = \frac{2}{3}$.
So,$x = \frac{2}{3}(x^2-y^2) \implies 3x = 2x^2 - 2y^2$ (This does not match options,re-evaluating).
Correcting the integration: $\int \frac{2v}{1-v^2} dv = \int \frac{dx}{x} \implies -\ln|1-v^2| = \ln|x| + \ln|k| \implies \ln|\frac{1}{1-v^2}| = \ln|kx| \implies \frac{1}{1-v^2} = kx$.
Using $(2, 1)$: $\frac{1}{1-(1/4)} = k(2) \implies \frac{1}{3/4} = 2k \implies \frac{4}{3} = 2k \implies k = \frac{2}{3}$.
$\frac{x^2}{x^2-y^2} = \frac{2}{3}x \implies 3x = 2(x^2-y^2) \implies 2y^2 = 2x^2 - 3x$. The provided options were incorrect; the correct curve is $2y^2 = 2x^2 - 3x$.

The slope of the tangent at $(x, y)$ to the curve passing through $(2, 1)$ is $\frac{x^2+y^2}{2xy}$. Find the equation of the curve.

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