JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 19)
The difference between degree and order of a differential equation that represents the family of curves given by $${y^2} = a\left( {x + {{\sqrt a } \over 2}} \right)$$, a > 0 is _________.
Answer
2
Explanation
$${y^2} = a\left( {x + {{\sqrt a } \over 2}} \right)$$
Differentiating both sides, we get
$$2yy' = a$$
$${y^2} = 2yy'\left( {x + {{\sqrt {2yy'} } \over 2}} \right)$$
$$y = 2y'\left( {x + {{\sqrt {yy'} } \over {\sqrt 2 }}} \right)$$
$$y - 2xy' = \sqrt 2 y'\sqrt {yy'} $$
$${\left( {y - 2x{{dy} \over {dx}}} \right)^2} = 2y{\left( {{{dy} \over {dx}}} \right)^3}$$
D = 3 & O = 1
$$ \therefore $$ D $$-$$ O = 3 $$-$$ 1 = 2
Differentiating both sides, we get
$$2yy' = a$$
$${y^2} = 2yy'\left( {x + {{\sqrt {2yy'} } \over 2}} \right)$$
$$y = 2y'\left( {x + {{\sqrt {yy'} } \over {\sqrt 2 }}} \right)$$
$$y - 2xy' = \sqrt 2 y'\sqrt {yy'} $$
$${\left( {y - 2x{{dy} \over {dx}}} \right)^2} = 2y{\left( {{{dy} \over {dx}}} \right)^3}$$
D = 3 & O = 1
$$ \therefore $$ D $$-$$ O = 3 $$-$$ 1 = 2
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