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 $Py"+Q{y}^{\prime }+1=0$, where P, Q are functions of x, y and y' (here $y^{\prime }=\frac{dy}{dx},y"=\frac{d^{2}y}{d{x}^2})$, 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)

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

We have ,
$(x-h{)}^2+(y-h{)}^2=r^2$
Differentiating and dividing by 2,
$x-h+y{y}^{\prime }-h{y}^{\prime }=0$
$\Rightarrow h=\frac{x+y{y}^{\prime }}{1+y^{\prime }}$
$h(1+y^{\prime })=x+y{y}^{\prime }$

Again difference wrt x,

$h{y}^{\prime \prime }=1+y{y}^{\prime \prime }+y^{\prime 2}$

Putting value of h,

$\left(\frac{x+y{y}^{\prime }}{1+y^{\prime }}\right)y^{\prime }=1+y{y}^{\prime \prime }+y^{\prime 2}$

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

$\therefore x{y}^{\prime \prime }+y{y}^{\prime }{y}^{\prime \prime }=1+y{y}^{\prime \prime }+y^{\prime 2}+y^{\prime }+y{y}^{\prime \prime }{y}^{\prime }+y^{\prime 3}$

$\therefore x{y}^{\prime \prime }=1+y{y}^{\prime \prime }+y^{\prime 2}+y^{\prime }+y^{\prime 3}$

$1+y{y}^{\prime \prime }+y^{\prime 2}+y^{\prime 3}+y{y}^{\prime \prime }-x{y}^{\prime \prime }=0$

$1+y^{\prime }+y^{\prime 2}+y^{\prime 3}+y{y}^{\prime \prime }-x{y}^{\prime \prime }=0$

$\therefore 1+y^{\prime }(1+y^{\prime }+y^{\prime 2})+y^{\prime \prime }(y-x)=0$

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

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