Question

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 $Py" + Qy' + 1 = 0$, where $P$, $Q$ are functions of $x, y$ and $y'$ (here $y' = \frac{dy}{dx}, \,y'' = \frac{d^{2}y}{dx^{2}}$), then which of the following statements is (are) true ?

• A. $P = y + x$
• B. $P = y - x$
• C. $P + Q = 1 - x + y + y' + (y')^2$
• D. $P - Q = x + y - y' - (y')^2$

Answer 1

Answer: (C)

$P + Q = 1 - x + y + y' + (y')^2$

Answer 2

Let the equation of circle is
$\left(x - a\right)^{2 }+ \left(y - a\right)^{2} = r^{2}$
$? x^{2} + y^{2 }- 2\, a\, x - 2\, a \,y + 2\, a^{ 2} - r^{2} = 0$
differentiate w.r.t. x
$? 2x + 2yy' - 2 \,a - 2\, a\, y' = 0$
$\Rightarrow \alpha = \frac{x+yy'}{1+y'}\quad\quad...\left(i\right)$
differentiate again w.r.t. x
$2 + 2\left(y'\right)^{2} + 2yy - 2 \,a\, y = 0$
$\Rightarrow \alpha = \frac{1+\left(y'\right)^{2}+yy''}{y''}\quad\quad...\left(ii\right)$
from $\left(i\right) \& \left(ii\right)$
$xy + yy'y = 1 + \left(y'\right)^{2} + yy + y' + \left(y'\right)^{3} + yy'y''$
$? \left(y - x\right) y + y' \left[y' + 1 + \left(y'\right)^{2}\right] + 1 = 0$
$P = y - x$
$Q = y' + 1 + \left(y'\right)^{2}$