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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.

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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}

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