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Mathematics Question on General and Particular Solutions of a Differential Equation

The differential equation of the system of all circles of radius rr in the xyxy plane is

A

[1+(dydx)3]2=r2(d2ydx2)2 \left[1+\left(\frac{dy}{dx}\right)^{^3}\right]^{^{^2}}=r^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{^2}

B

[1+(dydx)3]2=r2(d2ydx2)3 \left[1+\left(\frac{dy}{dx}\right)^{^3}\right]^{^{^2}}=r^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{^3}

C

[1+(dydx)2]3=r2(d2ydx2)2 \left[1+\left(\frac{dy}{dx}\right)^{^2}\right]^{^{^3}}=r^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{^2}

D

[1+(dydx)2]3=r2(d2ydx2)3 \left[1+\left(\frac{dy}{dx}\right)^{^2}\right]^{^{^3}}=r^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{^3}

Answer

[1+(dydx)2]3=r2(d2ydx2)2 \left[1+\left(\frac{dy}{dx}\right)^{^2}\right]^{^{^3}}=r^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{^2}

Explanation

Solution

The equation of the family of circles of radius r is
(x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2 \quad \quad \quad ...(1)
Where a & b are arbitrary constants.
Since equation (1) contains two arbitrary constants, we differentiate it two times w.r.t x & the differential equation will be of second order.
Differentiating (1) w.r.t. x, we get
2(x−a)+2(y−b)dydx=02\left(x-a\right)+2\left(y-b\right) \frac{dy}{dx}=0
⇒(x−a)+2(y−b)dydx=0...(2)\Rightarrow \left(x-a\right)+2\left(y-b\right) \frac{dy}{dx}=0\quad\quad\quad\quad...\left(2\right)
Differentiating (2) w.r.t. x, we get
1+(y−b)d2ydx2+(dydx)2=0...(3)1+\left(y-b\right) \frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{^2}=0\quad\quad\quad\quad...\left(3\right)
⇒(y−b)=−1+(dydx)2d2ydx2...(4)\Rightarrow \left(y-b\right)=-\frac{1+\left(\frac{dy}{dx}\right)^{^2}}{\frac{d^{2}y}{dx^{2}}}\quad\quad\quad\quad\quad...\left(4\right)
On putting the value of (y - b) in equation(2), we get
x−a=[1+(dydx)2]dydxd2ydx2...(5)x-a=\frac{\left[1+\left(\frac{dy}{dx}\right)^{^2}\right] \frac{dy}{dx}}{\frac{d^{2}y}{dx^{2}}}\quad\quad\quad\quad\quad...\left(5\right)
Substituting the values of (x - a) &(x - b)in (1), we get
[1+(dydx)2]2(dydx)2(d2ydx2)2+[1+(dydx)2]2(d2ydx2)2=r2\frac{\left[1+\left(\frac{dy}{dx}\right)^{^2}\right]^{^{^2}}\left(\frac{dy}{dx}\right)^{^2}}{\left(\frac{d^{2}y}{dx^{2}}\right)^{^2}}+\frac{\left[1+\left(\frac{dy}{dx}\right)^{^2}\right]^{^{^2}}}{\left(\frac{d^{2}y}{dx^{2}}\right)^{^2}}=r^{2}
⇒[1+(dydx)2]3=r2(d2ydx2)2\Rightarrow \left[1+\left(\frac{dy}{dx}\right)^{^2}\right]^{^{^3}}=r^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{^2}