Differential Equations 1 Question 13

13.

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 $P y^{\prime}+Q y^{\prime}+1=0$, where $P, Q$ are the 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 statement(s) is/are true? (2015 Adv.)

(a) $P=y+x$

(b) $P=y-x$

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

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

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Answer:

Correct Answer: 13. (b,c)

Solution:

  1. Since, centre lies on $y=x$.

$\therefore$ Equation of circle is

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

On differentiating, we get

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

$\Rightarrow x+y y^{\prime}-a-a y^{\prime} =0 $

$\Rightarrow a =\frac{x+y y^{\prime}}{1+y^{\prime}}$

Again differentiating, we get

$0=\frac{\left(1+y^{\prime}\right)\left[1+y y^{\prime}+\left(y^{\prime}\right)^{2}\right]-\left(x+y y^{\prime}\right) \cdot\left(y^{\prime \prime}\right)}{\left(1+y^{\prime}\right)^{2}} $

$\Rightarrow \quad\left(1+y^{\prime}\right)\left[1+\left(y^{\prime}\right)^{2}+y y^{\prime \prime}\right]-\left(x+y y^{\prime}\right)\left(y^{\prime \prime}\right) =0 $

$\Rightarrow \quad 1+y^{\prime}\left[\left(y^{\prime}\right)^{2}+y^{\prime}+1\right]+y^{\prime \prime}(y-x) =0$

On comparing with $P y^{\prime \prime}+Q y^{\prime}+1=0$, we get

$ \begin{array}{ll} & P=y-x \\ \text { and } \quad Q & =\left(y^{\prime}\right)^{2}+y^{\prime}+1 \end{array} $



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