The curve which passes through the point (sqr | vedclass

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(D) Given the differential equation: $\frac{dy}{dx} = \frac{2x}{3y}$.Separating the variables,we get: $3y \, dy = 2x \, dx$.Integrating both sides: $\

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a hyperbola

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(D) Given the differential equation: $\frac{dy}{dx} = \frac{2x}{3y}$.Separating the variables,we get: $3y \, dy = 2x \, dx$.Integrating both sides: $\int 3y \, dy = \int 2x \, dx$.This gives: $\frac{3y^2}{2} = x^2 + C_1$,or $3y^2 = 2x^2 + C$.Rearranging the terms: $2x^2 - 3y^2 = -C$,which is of the form $Ax^2 - By^2 = K$.Since the curve passes through $(\sqrt{2}, 1)$,substitute these values: $2(\sqrt{2})^2 - 3(1)^2 = K \implies 2(2) - 3 = K \implies K = 1$.The equation of the curve is $2x^2 - 3y^2 = 1$.This equation represents a hyperbola.

The curve which passes through the point $(\sqrt{2}, 1)$ and satisfies the differential equation $\frac{dy}{dx} = \frac{2x}{3y}$ represents:

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