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Chapter 13: Problem 1130

The differential equation of all circles passing through the origin and having their centers on the y-axis is: OR The differential equation for the family of curves \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{ay}=0\), where a is an arbitrary constant is: (A) \(\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) \mathrm{y}^{1}=2 \mathrm{xy}\) (B) \(2\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) \mathrm{y}^{1}=\mathrm{xy}\) (C) \(2\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) \mathrm{y}^{1}=\mathrm{xy}\) (D) \(\left(x^{2}+y^{2}\right) y^{1}=2 x y\)

Short Answer

Expert verified
The short answer is: (A) \(\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) \mathrm{y}^{1}=2 \mathrm{xy}\).

Step by step solution

01

Write the equation of the circle on the y-axis

The equation of a circle with its center on the y-axis can be represented as: \[ x^2 + (y-a)^2 = a^2 \]
02

Expand the equation

Expand the equation to make it easier to differentiate: \[ x^2 + y^2 - 2ay + a^2 = a^2 \] Since it must pass through origin: \[x^2 +y^2=2ay\]
03

Implicit differentiation

Implicitly differentiate both sides of the equation with respect to x: \[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 2a \frac{d}{dx}(y) \] Applying the derivatives: \[ 2x + 2y\frac{dy}{dx} = 2a\frac{dy}{dx} \]
04

Eliminate the arbitrary constant (a) from the equation

Divide both sides of the equation by 2: \[ x + y\frac{dy}{dx} = a\frac{dy}{dx} \] Now, we need to eliminate the arbitrary constant (a). We can use the equation from Step 2, \(x^2 + y^2 = 2ay\), to eliminate a: \[ a = \frac{x^2 + y^2}{2y} \] Now, substitute this into our differential equation from Step 3: \[ x + y\frac{dy}{dx} = \frac{x^2 + y^2}{2y} \frac{dy}{dx} \]
05

Solve for the differential equation

Multiply both sides of the equation by \(2y\) and simplify: \[ 2xy + 2y^2\frac{dy}{dx} = (x^2 + y^2)\frac{dy}{dx} \] Rearrange the equation to have all terms on one side: \[ (x^2 + y^2)\frac{dy}{dx} - 2y^2\frac{dy}{dx} = 2xy \] Factor out the common term \(\frac{dy}{dx}\): \[ \frac{dy}{dx}(x^2 + y^2 - 2y^2) = 2xy \] The resulting differential equation is: \[ \frac{dy}{dx}(x^2 - y^2) = 2xy \] Comparing the above equation with the options given, we find that the answer is (A) \(\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) \mathrm{y}^{1}=2 \mathrm{xy}\).

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Most popular questions from this chapter

Solution of the differential equation \(\cos \mathrm{x} \cdot \mathrm{dy}\) \(=y(\sin x-y) d x, 0

Solution of \((\mathrm{dy} / \mathrm{dx})=1+\mathrm{x}+\mathrm{y}^{2}+\mathrm{xy}^{2}, \mathrm{y}(0)=0\) is: (A) \(y=\tan \left(c+x+x^{2}\right)\) (B) \(\mathrm{y}=\tan \left[\mathrm{x}+\left(\mathrm{x}^{2} / 2\right)\right]\) (C) \(\mathrm{y}^{2}=\exp \left[\mathrm{x}+\left(\mathrm{x}^{2} / 2\right)\right]\) (D) \(\mathrm{y}^{2}=1+\mathrm{c} \cdot \exp \left[\mathrm{x}+\left(\mathrm{x}^{2} / 2\right)\right]\)

The differential equation of family of parabolas with focus at origin and \(\mathrm{x}\) -axis as axis is: (A) \(\mathrm{y}(\mathrm{dy} / \mathrm{dx})^{2}-2 \mathrm{x}(\mathrm{dy} / \mathrm{dx})=\mathrm{y}\) (B) \(\mathrm{y}(\mathrm{dy} / \mathrm{dx})^{2}+2 \mathrm{xy}(\mathrm{dy} / \mathrm{dx})=\mathrm{y}\) (C) \(\mathrm{y}(\mathrm{dy} / \mathrm{dx})^{2}-2 \mathrm{xy}(\mathrm{dy} / \mathrm{dx})=\mathrm{y}\) (D) \(\mathrm{y}(\mathrm{dy} / \mathrm{dx})^{2}+2 \mathrm{x}(\mathrm{dy} / \mathrm{dx})=\mathrm{y}\)

The slope of the tangent at \((x, y)\) to a curve passing through \([1,(\pi / 4)]\) is given by \((\mathrm{y} / \mathrm{x})-\cos ^{2}(\mathrm{y} / \mathrm{x})\), then the equation of the curve is: (A) \(y=\tan ^{-1}[\log (\mathrm{e} / \mathrm{x})]\) (B) \(\mathrm{y}=\mathrm{x} \cdot \tan ^{-1}[\log (\mathrm{e} / \mathrm{x})]\) (C) \(\mathrm{y}=\mathrm{xtan}^{-1}(\mathrm{x} / \mathrm{e})\) (D) none of these

If \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) and \([(2+\sin \mathrm{x}) /(\mathrm{y}+1)](\mathrm{dy} / \mathrm{dx})=-\cos \mathrm{x}, \mathrm{y}(0)=1\) then y \((\pi / 2)\) equal: (A) \([(-1) / 3]\) (B) \((1 / 3)\) (C) \((2 / 3)\) (D) 1
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