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Please solve RD Sharma class 12 chapter Differential Equation exercise 21.2 question 16 subquestion (v) maths textbook solution

Answers (1)

Answer:

 The required differential equation is

y^{2}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-y^{2}=1

Hint:

 Differentiating equation (i) with respect to x

Given:

 (x-a)^{2}-y^{2}=1

Solution:

The equation of family of curves is

(x-a)^{2}-y^{2}=1 \qquad \qquad \dots(i)

Where a is a parameter

Differentiating equation (i) with respect to x, we get

\begin{aligned} &2(x-a)-2y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &(x-a)-y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &\sqrt{1+y^{2}}=y\frac{\mathrm{d} y}{\mathrm{d} x} \qquad \qquad [U\! sing (i)]\\ &(1+y^{2})=y^{2}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \end{aligned}

 The required equation is

y^{2}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-y^{2}=1

Posted by

Gurleen Kaur

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