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The differential equation (dy)/(dx)=(2x)...

The differential equation `(dy)/(dx)=(2x)/(3y)` represents a family of hyperbolas (except when it represents a pair of lines) with eccentricity. (a)`sqrt(3/5)` (b) `sqrt(5/3)` (c)`sqrt(2/5)` (d) `sqrt(5/2)`

A

`sqrt(3//5)`

B

`sqrt(5//3)`

C

`sqrt(2//5)`

D

`sqrt(5//2)`

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To solve the differential equation \(\frac{dy}{dx} = \frac{2x}{3y}\) and find the eccentricity of the family of hyperbolas it represents, we can follow these steps: ### Step 1: Separate the Variables We start with the given differential equation: \[ \frac{dy}{dx} = \frac{2x}{3y} \] We can separate the variables by multiplying both sides by \(3y\) and \(dx\): \[ 3y \, dy = 2x \, dx \] ### Step 2: Integrate Both Sides Now we integrate both sides: \[ \int 3y \, dy = \int 2x \, dx \] The left side integrates to: \[ \frac{3y^2}{2} + C_1 \] And the right side integrates to: \[ x^2 + C_2 \] Combining these, we have: \[ \frac{3y^2}{2} = x^2 + C \] where \(C = C_2 - C_1\). ### Step 3: Rearranging the Equation Rearranging the equation gives us: \[ 3y^2 = 2x^2 + 2C \] Dividing through by \(2C\) (assuming \(C \neq 0\)): \[ \frac{y^2}{\frac{2C}{3}} - \frac{x^2}{C} = 1 \] This is in the standard form of a hyperbola: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \] where \(a^2 = \frac{2C}{3}\) and \(b^2 = C\). ### Step 4: Finding the Eccentricity The eccentricity \(e\) of a hyperbola is given by: \[ e = \sqrt{1 + \frac{b^2}{a^2}} \] Substituting the values of \(a^2\) and \(b^2\): \[ e = \sqrt{1 + \frac{C}{\frac{2C}{3}}} \] This simplifies to: \[ e = \sqrt{1 + \frac{3}{2}} = \sqrt{\frac{5}{2}} \] ### Step 5: Consider the Case for \(C < 0\) If \(C < 0\), we can repeat the steps: The equation becomes: \[ \frac{y^2}{\frac{2(-C)}{3}} - \frac{x^2}{-C} = 1 \] This gives: \[ e = \sqrt{1 + \frac{-C}{\frac{2(-C)}{3}}} = \sqrt{1 + \frac{3}{2}} = \sqrt{\frac{5}{3}} \] ### Conclusion Thus, the eccentricities of the hyperbolas represented by the differential equation are: - For \(C > 0\), \(e = \sqrt{\frac{5}{2}}\) - For \(C < 0\), \(e = \sqrt{\frac{5}{3}}\) The correct options are: (b) \(\sqrt{\frac{5}{3}}\) and (d) \(\sqrt{\frac{5}{2}}\).

To solve the differential equation \(\frac{dy}{dx} = \frac{2x}{3y}\) and find the eccentricity of the family of hyperbolas it represents, we can follow these steps: ### Step 1: Separate the Variables We start with the given differential equation: \[ \frac{dy}{dx} = \frac{2x}{3y} \] We can separate the variables by multiplying both sides by \(3y\) and \(dx\): ...
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The differential equation (dx)/(dy)=(3y)/(2x) represents a family of hyperbolas (except when it represents a pair of lines) with eccentricity. sqrt(3/5) (b) sqrt(5/3) sqrt(2/5) (d) sqrt(5/2)

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