#### Please solve RD Sharma class 12 chapter Differential Equation exercise 21.2 question 19 maths textbook solution

The required differential equation is

$\left [ (x+y)^{2}+1 \right ]\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$

Hint:

The equation belongs to the family of a circle

$(x-b)^{2}+(y-k)^{2}=r^{2}$

Given:

The family of circles in the second quadrant and touching the co-ordinate axes

Solution:

Let c denotes the family of circles in the second quadrant and touching the coordinate axes and let(-a,a) be coordinate of the centre of any member of this circle

Now, the equation representing this family of circle is

\begin{aligned} &(x+a)^{2}+(y-a)^{2}=a^{2} \qquad \qquad \dots (i) \\ &x^{2}+y^{2}+2ax-2ay+a^{2}=0\qquad \qquad \dots (ii) \end{aligned}

Differentiating (ii) with respect to x, we get

\begin{aligned} &2x+2y\frac{\mathrm{d} y}{\mathrm{d} x}+2a-2a\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &x+y\frac{\mathrm{d} y}{\mathrm{d} x}+a-a\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &x+y\frac{\mathrm{d} y}{\mathrm{d} x}=-a+a\frac{\mathrm{d} y}{\mathrm{d} x} \\ &a=\frac{\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )}{\frac{\mathrm{d} y}{\mathrm{d} x}-1} \end{aligned}

Substituting this value of a in (i), we get

\begin{aligned} &\left [ x+\frac{\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )}{\frac{\mathrm{d} y}{\mathrm{d} x}-1} \right ]^{2}+\left [ y-\frac{\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )}{\frac{\mathrm{d} y}{\mathrm{d} x}-1} \right ]^{2}=\left [\frac{\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )}{\frac{\mathrm{d} y}{\mathrm{d} x}-1} \right ]^{2} \end{aligned}

\begin{aligned} &\left [ x\left ( \frac{\mathrm{d} y}{\mathrm{d} x}-1 \right ) +\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right ) \right ]^{2}+\left [ y\left ( \frac{\mathrm{d} y}{\mathrm{d} x}-1 \right ) +\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right ) \right ]^{2}=\left [ x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right ]^{2} \end{aligned}

\begin{aligned} &(x+y)^{2}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )+(x+y)^{2}=\left [ (x+y)^{2} \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )\right ]^{2} \end{aligned}

$\left [ (x+y)^{2}+1 \right ]\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$

The required differential equation is

$\left [ (x+y)^{2}+1 \right ]\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}$