The differential equation representing the family of curves $y^{2} = 2c(x + \sqrt{c})$, where $c 0$ is a parameter is of order and degree as follows

Question

The differential equation representing the family of curves $y^{2} = 2c(x + \sqrt{c})$, where $c > 0$ is a parameter is of order and degree as follows

Dr. Abubacker · Accepted Answer

(b)
Given, $y^{2} = 2c(x + \sqrt{c})$
$\Rightarrow$ $2yy_{1} = 2c$
$\Rightarrow$ $c = yy_{1}$
$	herefore$ $y^{2} = 2yy_{1}(x + \sqrt{yy_{1}})$
$\Rightarrow$ $y^{2} - 2yy_{1}x = \sqrt{yy_{1}}.2yy_{1}$
$\Rightarrow$ $\left( y^{2} - 2yy_{1}x ight)^{2} = 4\left( yy_{1} ight)^{3}$
$	herefore$ The degree of above equation is 3 and order is 1.

Q. The differential equation representing the family of curves $y^{2} = 2c(x + \sqrt{c})$, where $c > 0$ is a parameter is of order and degree as follows

Explanation

Reviewed by Dr. Abubacker

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