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

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


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



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)


\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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