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

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

Answers (1)

Answer:

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}

Posted by

Gurleen Kaur

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