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

• A. $P=y+x$
• B. $P=y-x$
• C. $P+Q=1-x+y+y^{\prime }+(y^{\prime }{)}^2$
• D. $P-Q=x+y-y^{\prime }-(y^{\prime }{)}^2$

Answer 1

Answer: (B), (C)

$P+Q=1-x+y+y^{\prime }+(y^{\prime }{)}^2$

As the center of the circle lies on the line $y=x,$
So equation of the circle is
$(x-h{)}^2+(y-h{)}^2=r^2$
Now differentiate w.r.t. $x,$ we get
$2x-2h+2(y-h)y^{\prime }=0\Rightarrow 2x+2y{y}^{\prime }–2h–2h{y}^{\prime }=0\Rightarrow h=\frac{x+y{y}^{\prime }}{1+y^{\prime }}\cdots \left(1\right)$
Again differentiate w.r.t. $x,$ we get
$2+2(y^{\prime }{)}^2+2y{y}^{\prime \prime }–2h{y}^{\prime \prime }=0$
Now putting the value of $h,$ we get
$2+2(y^{\prime }{)}^2+2y{y}^{\prime \prime }–2\left(\frac{x+y{y}^{\prime }}{1+y^{\prime }}\right)y^{\prime \prime }=0\Rightarrow 1+y^{\prime }+(y^{\prime }{)}^2+(y^{\prime }{)}^3+y{y}^{\prime \prime }+y{y}^{\prime }{y}^{\prime \prime }–x{y}^{\prime \prime }–y{y}^{\prime }{y}^{\prime \prime }=0\Rightarrow (y–x)y^{\prime \prime }+(1+y^{\prime }+\left(y^{\prime }{)}^2\right)y^{\prime }+1=0\Rightarrow (P{y}^{\prime \prime }+Q{y}^{\prime }+1=0)\text{where}P=y-x,Q=1+y^{\prime }+(y^{\prime }{)}^2$

$P+Q=1-x+y+y^{\prime }+(y^{\prime }{)}^2$ and $P=y-x$ are the correct options.