The differential equation representing the family of curves \[{y^2} = 2c\left( {x + \sqrt c } \right),\] where c>0, is a parameter, is of order and degree as follows:
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Official solution
\[{y^2} = 2c\left( {x + \sqrt c } \right)\,\,\,\,\,\,\,\,\,\,...\left( i \right)\]
\[2yy' = 2c.1\,\,\,or\,\,\,yy' = c\,\,\,...\left( {ii} \right)\]
\[ \Rightarrow {y^2} = 2yy'\left( {x + \sqrt {yy'} } \right)\]
\[\left[ \, \right.\] On putting value of c from (ii) in (i) \[\left. \, \right]\]
On simplifying, we get
\[{\left( {y - 2xy'} \right)^2} = 4yy{'^3}\,\,\,\,\,\,\,\,\,...\left( {iii} \right)\]
Hence equation (iii) is of order 1 and degree 3.