#### Provide solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.2 question 7

$y'^{2}(x^{2}-2y^{2})-4xyy'-x^{2}=0$

Hint:

Differentiating the given equation with respect to x

Given:

$y^{2}-2ay+x^{2}=a^{2}$

Solution:

$y^{2}-2ay+x^{2}=a^{2}$

Differentiating with respect to x

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

Putting the value of a in the given equation, we get

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

put

\begin{aligned} \frac{\mathrm{d} y}{\mathrm{d} x}=y' \end{aligned}

\begin{aligned} &y^{2}-2a\left ( \frac{x+yy'}{y'} \right )y+x^{2}=\left ( \frac{x+yy'}{y'} \right )^{2} \\ &\frac{y^{2}y'-2(y^{2}y'+xy)+x^{2}y}{y'}=\frac{y^{2}y'^{2}+2xyy'+x^{2}}{y'^{2}} \end{aligned}

\begin{aligned} &y^{2}y'^{2}-2y^{2}y'^{2}-2xyy'+x^{2}y'^{2}-y^{2}y'-2xyy'-x^{2}=0 \\ &-4xyy'+x^{2}y'^{2}-x^{2}-2y^{2}y'^{2}=0 \\ &y'^{2}(x^{2}-2y^{2})-4xyy'-x^{2}=0 \end{aligned}