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  • Explain solution RD Sharma class 12 chapter Differential Equation exercise 21.2 question 16 subquestion (iv) maths

Explain solution RD Sharma class 12 chapter Differential Equation exercise 21.2 question 16 subquestion (iv) maths

Answers (1)

Answer:

 The required differential equation is

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

Hint:

 \text { Putting } \sqrt{1-x^{2}} \text { instead of } (y-b) \text { in }2x+2(y-b)\frac{\mathrm{d} y}{\mathrm{d} x}=0

Given:

 x^{2}+(y-b)^{2}=1

Solution:

The equation of family of curves is

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

Where b is a parameter

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

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

 The required equation is

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

Posted by

Gurleen Kaur

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