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. $(0.40)$ $=0, x^{2} \frac{d y}{d x}+1=0$
Solution:
- 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
$$ \begin{aligned} & & 2 x+2 y y^{\prime}-2 a-2 a y^{\prime} & =0 \\ \Rightarrow & & x+y y^{\prime}-a-a y^{\prime} & =0 \\ \Rightarrow & & a & =\frac{x+y y^{\prime}}{1+y^{\prime}} \end{aligned} $$
Again differentiating, we get
$$ \begin{aligned} & 0=\frac{\left(1+y^{\prime}\right)\left[1+y y^{\prime}+\left(y^{\prime}\right)^{2}\right]-\left(x+y y^{\prime}\right) \cdot\left(y^{\prime \prime}\right)}{\left(1+y^{\prime}\right)^{2}} \\ \Rightarrow \quad\left(1+y^{\prime}\right)\left[1+\left(y^{\prime}\right)^{2}+y y^{\prime \prime}\right]-\left(x+y y^{\prime}\right)\left(y^{\prime \prime}\right) & =0 \\ \Rightarrow \quad 1+y^{\prime}\left[\left(y^{\prime}\right)^{2}+y^{\prime}+1\right]+y^{\prime \prime}(y-x) & =0 \end{aligned} $$
On comparing with $P y^{\prime \prime}+Q y^{\prime}+1=0$, we get
$$ \begin{array}{ll} & P=y-x \\ \text { and } \quad Q & =\left(y^{\prime}\right)^{2}+y^{\prime}+1 \end{array} $$
