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Chapter 13: Problem 1125

The differential equation of family of circles of radius 'a' is: (A) \(a^{2} y_{2}=\left[1-y_{1}^{3}\right]^{2}\) (B) \(a^{2} y_{2}=\left[1-y_{1}^{2}\right]^{3}\) (C) \(a^{2}\left(y_{2}\right)^{2}=\left[1+y_{1}^{3}\right]^{2}\) (D) \(a^{2}\left(y_{2}\right)^{2}=\left[1+y_{1}^{2}\right]^{3}\)

Short Answer

Expert verified
The short answer is: The differential equation of the family of circles of radius 'a' is (D) \(a^2(y_2)^2 = (1 + y_1^2)^3\).

Step by step solution

01

Write the general equation of a circle

We will start with the equation of a circle given by \((x - h)^2 + (y - k)^2 = a^2\). Here, \((h, k)\) is the center and \(a\) is the radius.
02

Implicit differentiation

Now, we need to find the differential equation. We will differentiate the given equation with respect to \(y\). Keep in mind that when we differentiate with respect to \(y\), the terms involving \(y\) will also include a factor of \(y'\): \[\frac{d}{dy} [(x - h)^2 + (y - k)^2] = \frac{d}{dy} [a^2]\]
03

Evaluate the derivatives

Now we will evaluate the derivatives: \[\frac{d}{dy}[(x - h)^2] + \frac{d}{dy}[(y - k)^2] = 0\] Since \((x - h)^2\) does not have a \(y\) component, its derivative with respect to \(y\) is 0. \[0 + 2(y - k)(y' - 0) = 0\]
04

Simplify the equation

Now we will simplify the equation we got in Step 3: \[2(y - k)(y') = 0\] Since we want to find the equation in terms of \(y_1\) and \(y_2\), we can substitute \(y_1 = y - k\) and \(y_2 = y'\). The equation becomes: \[2(y_1)(y_2) = 0\] Now, we will substitute back the equation of a circle: \[(x - h)^2 + y_1^2 = a^2\] Since we have one equation and one unknown, we can solve for \(h\): \[h = x - \sqrt{a^2 - y_1^2}\] Now we will find the expression that represents the family of circles: \(a^2 y_2^2 = f(y_1)\)
05

Compare with given options

Now we will compare our expression for the family of circles with the given options: (A) \(a^2y_2 = (1 - y_1^3)^2\): This option doesn't match our result. (B) \(a^2y_2 = (1 - y_1^2)^3\): This option doesn't match our result. (C) \(a^2(y_2)^2 = (1 + y_1^3)^2\): This option doesn't match our result. (D) \(a^2(y_2)^2 = (1 + y_1^2)^3\): This option matches our result. So, the correct answer is (D) \(a^2(y_2)^2 = (1 + y_1^2)^3\).

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Most popular questions from this chapter

If the slope of tangent at \((\mathrm{x}, \mathrm{y})\) to the curve passing through \((2,1)\) is \(\left[\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) /(2 \mathrm{xy})\right]\). The equation of the curve is: (A) \(2\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)=6 \mathrm{y}\) (B) \(2\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)=3 \mathrm{x}\) (C) \(x\left(x^{2}+y^{2}\right)=10\) (D) \(x\left(x^{2}-y^{2}\right)=6\)

The particular solution of \(\left(1+\mathrm{y}^{2}\right) \mathrm{dx}+\left(\mathrm{x}-\mathrm{e}^{-(\tan )-1 \mathrm{y}}\right) \mathrm{dy}=0\) with initial condition \(\mathrm{y}(0)=0\) is: (A) \(x \cdot e^{(\tan )-1 x}=\cot ^{-1} x\) (B) \(x \cdot e^{(\tan )-1 y}=\tan ^{-1} y\) (C) \(x \cdot e^{(\tan )-1 y}=\cot ^{-1} y\) (D) \(x \cdot e^{(\cot )-1 y}=\tan ^{-1} y\)

If \(\mathrm{x}(\mathrm{dy} / \mathrm{dx})=\mathrm{y}(\log \mathrm{y}-\log \mathrm{x}+1)\), then the solution of the equation is: (A) \(x \log (\mathrm{y} / \mathrm{x})=\mathrm{cy}\) (B) \(\log (\mathrm{y} / \mathrm{x})=\mathrm{cx}\) (C) \(\log (\mathrm{x} / \mathrm{y})=\mathrm{cy}\) (D) \(\mathrm{y} \cdot \log (\mathrm{x} / \mathrm{y})=\mathrm{cx}\)

\(\underline{\text { Assertion - Reason Type Questions: }}\) Each question has four choices (a), (b), (c) and (d) out of which only one is correct. Write (a), (b), (c) and (d) according to the following rules. (a) Statement- 1 is True, Statement-2 is True, Statement- 2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement- 2 is True, Statement-2 is not a correct explanation for Statement-1. (c) Statement- 1 is True, Statement- 2 is False. (d) Statement- 1 is False, Statement- 2 is True. Statement \(-2:\) The differential equation \(\mathrm{y}^{\prime}=(\mathrm{y} / 2 \mathrm{x})\) is variable separable. Statement-1: Curve satisfying the differential equation \((\mathrm{dy} / \mathrm{dx})=(\mathrm{y} / 2 \mathrm{x})\) passing through \((2,1)\) is a parabola with Focus \([(1 / 4), \underline{0}]\). Statement- 2 : The differential equation \((\mathrm{dy} / \mathrm{dx})=(\mathrm{y} / 2 \mathrm{x})\) is variable separable.

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