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Chapter 13: Problem 1105

The order of the differential equation of family of circle touching a fixed straight line passing through origin is. (A) 2 (B) 3 (C) 4 (D) none of these

Short Answer

Expert verified
The order of the differential equation of the family of circles touching a fixed straight line passing through the origin is \(2\). Hence, the answer is (A) 2.

Step by step solution

01

Write the equation of a circle

A circle with center (h, k) and radius r is given by \((x-h)^2 + (y-k)^2 = r^2\). Now, let's derive the equation of the straight line passing through the origin.
02

Determine the equation of the straight line passing through the origin

Let the fixed straight line passing through the origin have a slope of m. The equation is given by \(y = mx\). Next, we will find the condition that the circle touches the straight line. This means that at some point on the circle, the line is tangent to the circle.
03

Find the condition for the line to touch the circle

The line touches the circle when the distance from the center of the circle to the line is equal to the radius, so we have \(r = \frac{|mk - h|}{\sqrt{1 + m^2}}\). Now, we will eliminate the arbitrary constants, (h, k) and r, from the circle equation.
04

Eliminate the arbitrary constants

Replace r in the circle equation with the expression in Step 3: \[(x - h)^2 + (y - k)^2 = \left(\frac{|mk - h|}{\sqrt{1 + m^2}}\right)^2\] We can use the equation of the line to replace k with k = y/mx in the above equation. After eliminating the constants and simplifying the equation, we will differentiate the resulting equation to obtain the differential equation.
05

Differentiate and find the order

Differentiate the equation obtained in Step 4 with respect to x once: \(2(x-h) + 2(y-k)\frac{dy}{dx} = 0\) Now differentiate again the latter equation once: \(\frac{d^2y}{dx^2} = \frac{1}{(k - y)^2(1 + m^2)}\) Since the equation now only contains second-order derivatives, the order of the differential equation is 2. Hence, the answer is (A) 2.

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Most popular questions from this chapter

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) \(\left(x^{2}+y^{2}\right) y^{1}=2 x y\)

The differential equation representing the family of curves \(\mathrm{y}^{2}=2 \mathrm{c}(\mathrm{x}+\mathrm{v} \mathrm{c})\), where \(\mathrm{c}\) is a positive parameter, is of order and degree as follows. (A) order 2, degree 1 (B) order 1 , degree 2 (C) order 2 , degree 2 (D) order 1 , degree 3

Solution of \(\left(d^{2} y / d x^{2}\right)=\log x\) is: (A) \(\mathrm{y}=(1 / 2) \mathrm{x}^{2} \log \mathrm{x}-(3 / 4) \mathrm{x}^{2}+\mathrm{c}_{1} \mathrm{x}+\mathrm{c}_{2}\) (B) \(\mathrm{y}=(1 / 2) \mathrm{x}^{2} \log \mathrm{x}+(3 / 4) \mathrm{x}^{2}+\mathrm{c}_{1} \mathrm{x}+\mathrm{c}_{2}\) (C) \(\mathrm{y}=(1 / 2) \mathrm{x}^{2} \log \mathrm{x}-(3 / 4) \mathrm{x}^{2}-\mathrm{c}_{1} \mathrm{x}+\mathrm{c}_{2}\) (D) None of these

The solution of the equation \((2 x+y+1) d x+(4 x+2 y-1) d y\) \(=0\) is: (A) \(\log |2 \mathrm{x}+\mathrm{y}-1|+\mathrm{x}+2 \mathrm{y}=\mathrm{c}\) (B) \(\log (2 \mathrm{x}+\mathrm{y}+1)+\mathrm{x}+2 \mathrm{y}=\mathrm{c}\) (C) \(\log |2 \mathrm{x}+\mathrm{y}-1|=\mathrm{c}+\mathrm{x}+\mathrm{y}\) (D) \(\log (4 x+2 y-1)=c+2 x+y\)

Solution of differential equation \((\mathrm{dy} / \mathrm{dx})+\mathrm{ay}=\mathrm{e}^{\mathrm{mx}}\) is: (A) \(y=e^{m x}+c \cdot e^{-a x}\) (B) \((\mathrm{a}+\mathrm{m}) \mathrm{y}=\mathrm{e}^{\mathrm{mx}}+\mathrm{c}\) (C) \((a+m) y=e^{m x}+c \cdot e^{-a x}\) (D) \(\mathrm{y} \cdot \mathrm{e}^{\mathrm{ax}}=\mathrm{m} \cdot \mathrm{e}^{\mathrm{mx}}+\mathrm{c}\)
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