1)

 Consider the family of all circles whose centres 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 the functions of x, y and y'   (here,    $y'=\frac{dy}{dx},y''=\frac{d^{2}y}{dx^{2}})$, then which of the following statement (s) 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:

Option B,C

Explanation:

 Since the centre lies on y=x.

      $\therefore$  Equation of circle is

                    $x^{2}+y^{2}-2ax-2ay+c=0$

  On differentiating , we get

          $2x+2yy'-2a-2ay'=0$

   $\Rightarrow$       $x+yy'-a-ay'=0$

  $\Rightarrow$    $a=\frac{x+yy'}{1+y'}$

 Again differentiating , we get

         0  $= \frac{(1+y')[1+yy'+(y')^{2}]-(x+yy').(y")}{(1+y')^{2}}$

$\Rightarrow$    $ (1+y')[1+(y')^{2}+yy"]- (x+yy')(y")=0$

$\Rightarrow$    $1+y'[(y')^{2}+y'+1]+y"(y-x)=0$

 On comparing  with   $Py"+Qy'+1=0,$ we get 

 P=y-x  and $Q= (y')^{2}+y'+1$