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  • Please solve RD Sharma class 12 chapter Differential Equation exercise 21.2 question 10 maths textbook solution

Please solve RD Sharma class 12 chapter Differential Equation exercise 21.2 question 10 maths textbook solution

Answers (1)

Answer:

 x^{2}-y^{2}+2xy\frac{\mathrm{d} y}{\mathrm{d} x}=0

Hint:

 Let the equation of the circle assuming R as radius and (a,b) as the centre of the circle is 

(x-a)^{2}+(y-b)^{2}=R^{2}

Given:

Circles passes through origin and their centre lie on x-axis. So the centre of the circle will be (a, 0)

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

Solution:

\begin{aligned} &(x-a)^{2}+y^{2}=a^{2} \\ &2(x-a)dx+2ydy=0 \\ &x-a=-y\frac{\mathrm{d} y}{\mathrm{d} x} \\ &a=x+y\frac{\mathrm{d} y}{\mathrm{d} x} \end{aligned}

Substituting these values in equation (i)

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

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Gurleen Kaur

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