JEE MAIN - Mathematics (2005 - No. 26)
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:
order $$1,$$ degree $$2$$
order $$1,$$ degree $$1$$
order $$1,$$ degree $$3$$
order $$2,$$ degree $$2$$
Explanation
$${y^2} = 2c\left( {x + \sqrt c } \right)\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
$$2yy' = 2c.1\,\,\,or\,\,\,yy' = c\,\,\,...\left( {ii} \right)$$
$$ \Rightarrow {y^2} = 2yy'\left( {x + \sqrt {yy'} } \right)$$
$$\left[ \, \right.$$ 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 $$3.$$
$$2yy' = 2c.1\,\,\,or\,\,\,yy' = c\,\,\,...\left( {ii} \right)$$
$$ \Rightarrow {y^2} = 2yy'\left( {x + \sqrt {yy'} } \right)$$
$$\left[ \, \right.$$ 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 $$3.$$
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