#### Provide solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.2 question 15 subquestion (i)

$2xy\frac{\mathrm{d} y}{\mathrm{d} x}+4x^{2}-y^{2}=0$

Hint:

Using chain rule and putting the value of a in (i)

Given:

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

Solution:

\begin{aligned} &2(2x+a)2+2y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &2(2x+a)+y\frac{\mathrm{d} y}{\mathrm{d} x}=0\\ &(2x+a)=-\frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x}\\ &a=-\frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x}-2x \end{aligned}

\begin{aligned} &\left ( \frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+y^{2}=\left ( \frac{y}{2}\frac{\mathrm{d} y}{\mathrm{d} x}+2x \right )^{2}\\ &\left ( \frac{1}{2}y\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+y^{2}=\left ( \frac{y}{2}\frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+4x^{2}+2\frac{y}{2}.\frac{\mathrm{d} y}{\mathrm{d} x}.2x\\ &y^{2}-4x^{2}-2xy\frac{\mathrm{d} y}{\mathrm{d} x}=0 \end{aligned}

$2xy\frac{\mathrm{d} y}{\mathrm{d} x}+4x^{2}-y^{2}=0$