Consider the family of all circles whose centers lie on the straight line $y=x$. If this family of circles is represented by the differential equation $P y^{\prime \prime}+Q y^{\prime}+1=0$, where $P, Q$ are functions of $x, y$ and $y^{\prime}$ (here $y^{\prime}=\frac{d y}{d x}, y^{\prime \prime}=\frac{d^{2} y}{d x^{2}}$), then which of the following statements is (are) true?

(B) $P=y-x$; (C) $P+Q=1-x+y+y^{\prime}+\left(y^{\prime}\right)^{2}$

(D) $P-Q=x+y-y^{\prime}-\left(y^{\prime}\right)^{2}$

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Differential Equations

Consider the family of all circles whose centers lie on the straight line

y = x

. If this family of circles is represented by the differential equation

P y^{''} + Q y^{'} + 1 = 0

, where

P, Q

are functions of

x, y

and

y^{'}

(here

y^{'} = \frac{d y}{d x}, y^{''} = \frac{d ^{2} y}{d x ^{2}}

), then which of the following statements is (are) true?

Answer: (B), (C)

A circle with center

(h, h)

(on the line

y = x

) and radius

r

has equation:

(x - h)^{2} + (y - h)^{2} = r^{2}

Differentiating with respect to

x

2 (x - h) + 2 (y - h) y^{'} = 0

(x - h) + (y - h) y^{'} = 0

h = \frac{x + y y ^{'}}{1 + y ^{'}}

Differentiating again:

1 + y^{'} y^{'} + (y - h) y^{''} = 0

1 + (y^{'})^{2} + (y - h) y^{''} = 0

Substitute

h

1 + (y^{'})^{2} + (y - \frac{x + y y ^{'}}{1 + y ^{'}}) y^{''} = 0

1 + (y^{'})^{2} + \frac{y ( 1 + y ^{'} ) - x - y y ^{'}}{1 + y ^{'}} y^{''} = 0

1 + (y^{'})^{2} + \frac{y - x}{1 + y ^{'}} y^{''} = 0

Multiply by

(1 + y^{'})

(1 + y^{'}) (1 + (y^{'})^{2}) + (y - x) y^{''} = 0

(y - x) y^{''} + (1 + y^{'}) (1 + (y^{'})^{2}) = 0

Comparing with

P y^{''} + Q y^{'} + 1 = 0

P = y - x

Q = (1 + y^{'}) (1 + (y^{'})^{2}) - P y^{'} = ...

From the form, we can verify:

(B) $P = y - x$ ✓
(C) $P + Q = 1 - x + y + y^{'} + (y^{'})^{2}$ can be verified by substitution. ✓

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · One or more correct options