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 (integer)
2
Solution
${y^2} = a\left( {x + {{\sqrt a } \over 2}} \right)$
<br><br>Differentiating both sides, we get
<br><br>$2yy' = a$<br><br>${y^2} = 2yy'\left( {x + {{\sqrt {2yy'} } \over 2}} \right)$<br><br>$y = 2y'\left( {x + {{\sqrt {yy'} } \over {\sqrt 2 }}} \right)$<br><br>$y - 2xy' = \sqrt 2 y'\sqrt {yy'}$<br><br>$${\left( {y - 2x{{dy} \over {dx}}} \right)^2} = 2y{\left( {{{dy} \over {dx}}} \right)^3}$$<br><br>D = 3 & O = 1<br><br>$\therefore$ D $-$ O = 3 $-$ 1 = 2
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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