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Question: The differential equation whose solution is $(x - h)^2 + (y - k)^2 = a^2$ (a is given constant) is :...

The differential equation whose solution is (x−h)2+(y−k)2=a2(x - h)^2 + (y - k)^2 = a^2 (a is given constant) is :

A

{1+(dydx)2}3=a2d2ydx2\left\{ 1 + \left( \frac{dy}{dx} \right)^2 \right\}^3 = a^2 \frac{d^2y}{dx^2}

B

{1+(dydx)2}3=a2(d2ydx2)2\left\{ 1 + \left( \frac{dy}{dx} \right)^2 \right\}^3 = a^2 \left( \frac{d^2y}{dx^2} \right)^2

C

{1+(dydx)2}3=a2(d2ydx2)2\left\{ 1 + \left( \frac{dy}{dx} \right)^2 \right\}^3 = a^2 \left( \frac{d^2y}{dx^2} \right)^2

D

None of these

Answer

{1+(dydx)2}3=a2(d2ydx2)2\left\{ 1 + \left( \frac{dy}{dx} \right)^2 \right\}^3 = a^2 \left( \frac{d^2y}{dx^2} \right)^2

Explanation

Solution

The given equation of the family of curves is:

(x−h)2+(y−k)2=a2…(1)(x - h)^2 + (y - k)^2 = a^2 \quad \ldots(1)

Here, hh and kk are arbitrary constants, and aa is a given constant. Since there are two arbitrary constants (hh and kk), we need to differentiate the equation twice to eliminate them.

Step 1: Differentiate equation (1) with respect to xx.

Applying the chain rule:

2(x−h)⋅ddx(x−h)+2(y−k)⋅ddx(y−k)=ddx(a2)2(x - h) \cdot \frac{d}{dx}(x - h) + 2(y - k) \cdot \frac{d}{dx}(y - k) = \frac{d}{dx}(a^2)

2(x−h)(1)+2(y−k)(dydx)=02(x - h)(1) + 2(y - k)\left(\frac{dy}{dx}\right) = 0

Divide by 2:

(x−h)+(y−k)y′=0…(2)(x - h) + (y - k)y' = 0 \quad \ldots(2)

(where y′=dydxy' = \frac{dy}{dx})

Step 2: Differentiate equation (2) with respect to xx.

Applying the product rule for the second term:

ddx(x−h)+ddx((y−k)y′)=ddx(0)\frac{d}{dx}(x - h) + \frac{d}{dx}((y - k)y') = \frac{d}{dx}(0)

1+((y−k)ddx(y′)+y′ddx(y−k))=01 + \left((y - k) \frac{d}{dx}(y') + y' \frac{d}{dx}(y - k)\right) = 0

1+(y−k)y′′+y′(y′)=01 + (y - k)y'' + y'(y') = 0

1+(y−k)y′′+(y′)2=0…(3)1 + (y - k)y'' + (y')^2 = 0 \quad \ldots(3)

(where y′′=d2ydx2y'' = \frac{d^2y}{dx^2})

Step 3: Express (y−k)(y - k) and (x−h)(x - h) in terms of y′y' and y′′y''.

From equation (3), we can isolate (y−k)(y - k):

(y−k)y′′=−(1+(y′)2)(y - k)y'' = -(1 + (y')^2)

(y−k)=−1+(y′)2y′′…(4)(y - k) = -\frac{1 + (y')^2}{y''} \quad \ldots(4)

Now substitute this expression for (y−k)(y - k) into equation (2) to find (x−h)(x - h):

(x−h)+(−1+(y′)2y′′)y′=0(x - h) + \left(-\frac{1 + (y')^2}{y''}\right)y' = 0

(x−h)=y′(1+(y′)2)y′′…(5)(x - h) = \frac{y'(1 + (y')^2)}{y''} \quad \ldots(5)

Step 4: Substitute the expressions for (x−h)(x - h) and (y−k)(y - k) back into the original equation (1).

Substitute (4) and (5) into (1):

(y′(1+(y′)2)y′′)2+(−1+(y′)2y′′)2=a2\left(\frac{y'(1 + (y')^2)}{y''}\right)^2 + \left(-\frac{1 + (y')^2}{y''}\right)^2 = a^2

(y′)2(1+(y′)2)2(y′′)2+(1+(y′)2)2(y′′)2=a2\frac{(y')^2(1 + (y')^2)^2}{(y'')^2} + \frac{(1 + (y')^2)^2}{(y'')^2} = a^2

Since the denominators are the same, combine the numerators:

(y′)2(1+(y′)2)2+(1+(y′)2)2(y′′)2=a2\frac{(y')^2(1 + (y')^2)^2 + (1 + (y')^2)^2}{(y'')^2} = a^2

Factor out (1+(y′)2)2(1 + (y')^2)^2 from the numerator:

(1+(y′)2)2((y′)2+1)(y′′)2=a2\frac{(1 + (y')^2)^2 \left((y')^2 + 1\right)}{(y'')^2} = a^2

(1+(y′)2)3(y′′)2=a2\frac{(1 + (y')^2)^3}{(y'')^2} = a^2

Step 5: Rearrange the equation to match the options.

(1+(y′)2)3=a2(y′′)2(1 + (y')^2)^3 = a^2 (y'')^2

Substitute back y′=dydxy' = \frac{dy}{dx} and y′′=d2ydx2y'' = \frac{d^2y}{dx^2}:

{1+(dydx)2}3=a2(d2ydx2)2\left\{ 1 + \left( \frac{dy}{dx} \right)^2 \right\}^3 = a^2 \left( \frac{d^2y}{dx^2} \right)^2

This matches option (B). Note that option (C) is identical to option (B), which might be a typo in the question.