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

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Question

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2vx^2} = \frac{1 + v^2}{2v}$.$x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}$.Separating variables: $\frac{2v}{1 - v^2} dv = \frac{dx}{x}$.Integrating both sides: $-\ln|1 - v^2| = \ln|x| + \ln|c|$.$-\ln|1 - \frac{y^2}{x^2}| = \ln|xc| \implies \ln|\frac{x^2}{x^2 - y^2}| = \ln|xc|$.$\frac{x^2}{x^2 - y^2} = xc \implies x = c(x^2 - y^2)$.Since the curve passes through $(2, 1)$,we have $2 = c(2^2 - 1^2) = c(3) \implies c = \frac{2}{3}$.Thus,$x = \frac{2}{3}(x^2 - y^2)$,which simplifies to $3x = 2(x^2 - y^2)$.

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

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