#### Explain solution RD Sharma class 12 chapter Differential Equation exercise 21.2 question 9 maths

$(x^{2}-y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}-2xy=0$

Hint:

Let (0, k) be the centre of the circle with k as its centre and differentiating the equation of the circle

Given:

Circles pass through origin and their centre lie on y-axis

Solution:

\begin{aligned} &(x-0)^{2}+(y-k)^{2}=k^{2} \\ &x^{2}+(y-k)^{2}=k^{2} \\ &x^{2}+y^{2}-2ky=0 \\ &\frac{x^{2}+y^{2}}{2y}=k \end{aligned}

\begin{aligned} &\frac{2y\left ( 2x+2y\frac{\mathrm{d} y}{\mathrm{d} x} \right )-(x^{2}-y^{2})2\frac{\mathrm{d} y}{\mathrm{d} x}}{4y^{2}}=0 \\ &4y\left ( x+y\frac{\mathrm{d} y}{\mathrm{d} x} \right )-2(x^{2}+y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &4xy=4y^{2}\frac{\mathrm{d} y}{\mathrm{d} x}-2(x^{2}+y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}

\begin{aligned} &(4y^{2}-2x^{2}-2y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}+4xy=0 \\ &(2y^{2}-2x^{2})\frac{\mathrm{d} y}{\mathrm{d} x}+2xy=0 \\ &(y^{2}-x^{2})\frac{\mathrm{d} y}{\mathrm{d} x}+2xy=0 \\ &(x^{2}-y^{2})\frac{\mathrm{d} y}{\mathrm{d} x}-2xy=0 \end{aligned}

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