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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 17 Maths Textbook Solution.

Answer: $3y^{{}'}\left ( y^{{}'''} \right )^{2}=y^{{}'''}\left ( 1+\left ( y^{{}'} \right )^{2} \right )$

Hint: The number of constant is equal to the number of times we need to  differentiate.

Given: $x^{2}+y^{2}+2ax+2by+C=0$, find differential equation, not containing constants.

Solution: $x^{2}+y^{2}+2ax+2by+C=0$

Differentiate w.r.t x,

$2x + 2yy^{{}'} + 2a + 2by^{{}'} = 0$

Again, differentiate w.r.t x

$2 + 2(y^{{}'})^{2} + 2yy^{{}''} + 2by^{{}''} = 0$

$1 + (y^{{}'})^{2} + yy^{{}''} + by^{{}''} = 0$..taking 2 common

$b=\frac{-\left (1+\left ( y^{{}'} \right )^{2}+yy^{{}''} \right )}{y^{{}''}}$

We have,

$1+\left(y^{{}'}\right)^{2}+y y^{{}''}+b y^{{}''}=0$

Again differentiate,

$2 y^{{}'} y^{{}''}+y^{{}'} y^{{}''}+y y^{{}'''}+b y^{{}'''}=0$

On substituting values of (b),

\begin{aligned} &3 y^{{}'} y^{{}''}+y y^{{}'''}+\left(\frac{-\left(1+\left(y^{{}'}\right)^{2}+y y^{{}''}\right.}{y^{{}''}}\right) y^{{}'''}=0 \\ &3 y^{{}'}\left(y^{{}''}\right)^{2}+y y^{{}''} y^{{}'''}-y^{{}''}-\left(y^{{}'}\right)^{2} y^{{}'''}-y y^{{}'''} y^{{}''}=0 \\ &3 y^{{}'}\left(y^{{}''}\right)^{2}=y^{{}'''}\left(1+\left(y^{{}'}\right)^{2}\right) \end{aligned}