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The differential equation representing the family of circles touching y - axis at the origin is
- a)Linear and of first order
- b)Linear and of second order
- c)Non linear and of first order
- d)Non linear and of second order
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The differential equation representing the family of circles touching ...
Equation of circles touching y axis is given by
( x - a2) + y2 = a2 where a is parameter implies x2 + y2 - 2ax = 0
Differentiating w.r.t. x, we get
2x + 2yy' - 2a = 0
implies a = x + yy'
So, differential equation will be
x2 +y2- 2x(x + yy1) = 0
implies
So, differential equation is non-linear and of first order
( x - a2) + y2 = a2 where a is parameter implies x2 + y2 - 2ax = 0
Differentiating w.r.t. x, we get
2x + 2yy' - 2a = 0
implies a = x + yy'
So, differential equation will be
x2 +y2- 2x(x + yy1) = 0
implies
So, differential equation is non-linear and of first order
Most Upvoted Answer
The differential equation representing the family of circles touching ...
Differential equation representing the family of circles touching the y-axis at the origin is non-linear and of first order.
Explanation:
To find the differential equation representing the family of circles touching the y-axis at the origin, let's consider a generic circle with center (a, b) and radius r.
Equation of this circle can be written as:
(x - a)^2 + (y - b)^2 = r^2
Since the circle touches the y-axis at the origin, the center of the circle lies on the y-axis. Therefore, the x-coordinate of the center is 0.
Substituting x = 0 in the equation of the circle, we get:
(0 - a)^2 + (y - b)^2 = r^2
a^2 + (y - b)^2 = r^2
Since the circle touches the y-axis at the origin, the y-coordinate of the center is equal to the radius. Therefore, a = r.
Substituting a = r in the equation, we get:
r^2 + (y - b)^2 = r^2
(y - b)^2 = 0
Taking the square root of both sides, we get:
y - b = 0
y = b
So, the equation of the circle becomes:
x^2 + (y - b)^2 = b^2
Differentiating this equation with respect to x, we get:
2x + 2(y - b) * (dy/dx) = 0
Simplifying this equation, we get:
x + (y - b) * (dy/dx) = 0
This is a non-linear and first-order differential equation, which represents the family of circles touching the y-axis at the origin.
Therefore, the correct answer is option 'C' - Non-linear and of first order.
Explanation:
To find the differential equation representing the family of circles touching the y-axis at the origin, let's consider a generic circle with center (a, b) and radius r.
Equation of this circle can be written as:
(x - a)^2 + (y - b)^2 = r^2
Since the circle touches the y-axis at the origin, the center of the circle lies on the y-axis. Therefore, the x-coordinate of the center is 0.
Substituting x = 0 in the equation of the circle, we get:
(0 - a)^2 + (y - b)^2 = r^2
a^2 + (y - b)^2 = r^2
Since the circle touches the y-axis at the origin, the y-coordinate of the center is equal to the radius. Therefore, a = r.
Substituting a = r in the equation, we get:
r^2 + (y - b)^2 = r^2
(y - b)^2 = 0
Taking the square root of both sides, we get:
y - b = 0
y = b
So, the equation of the circle becomes:
x^2 + (y - b)^2 = b^2
Differentiating this equation with respect to x, we get:
2x + 2(y - b) * (dy/dx) = 0
Simplifying this equation, we get:
x + (y - b) * (dy/dx) = 0
This is a non-linear and first-order differential equation, which represents the family of circles touching the y-axis at the origin.
Therefore, the correct answer is option 'C' - Non-linear and of first order.
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The differential equation representing the family of circles touching y - axis at the origin isa)Linear and of first orderb)Linear and of second orderc)Non linear and of first orderd)Non linear and of second orderCorrect answer is option 'C'. Can you explain this answer?
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