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Differential Equations

Short Answer Type

61. Form the differential equation representing the family of curves y = a cos (x + b) where a and b are arbitrary constants.

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62. Find the differential equation of the family of curves y = A sin mx + B cos mx. where m is fixed, and A and B are arbitrary constants.

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63. Form the differential equation corresponding to y² = m (a² – x²) by eliminating m and a.

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64. Form the differential equation corresponding to y² = a (b – x) (b + x) by eliminating a and b.

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65. Form the differential equation of the family of curves represented by the equation (x – a)² + 2 y² = a², a being the parameter.

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Long Answer Type

66. Prove that x² – y² = c (x² + y² )² is the general solution of differential equation (x³ – 3 x y² ) dx = (y³ –3 x² y) dy . where c is a parameter.

Here, $straight x squared minus straight y squared space equals space straight c left parenthesis straight x squared plus straight y squared right parenthesis squared$ ...(1)
Differentiating w.r.t.x, we get,
$2 straight x minus 2 straight y dy over dx space equals space 2 straight c left parenthesis straight x squared plus straight y squared right parenthesis. space open parentheses 2 straight x plus 2 straight y dy over dx close parentheses$
$straight x minus straight y space dy over dx space equals space 2 straight c left parenthesis straight x squared plus straight y squared right parenthesis. space open parentheses straight x plus straight y dy over dx close parentheses$ ...(2)
Dividing (2) by (1), we get.
$fraction numerator straight x minus straight y begin display style dy over dx end style over denominator straight x squared minus straight y squared end fraction space equals space fraction numerator 2 open parentheses straight x plus straight y begin display style dy over dx end style close parentheses over denominator straight x squared plus straight y squared end fraction$
$therefore space space space straight x left parenthesis straight x squared plus straight y squared right parenthesis minus straight y left parenthesis straight x squared plus straight y squared right parenthesis space dy over dx space equals space 2 straight x left parenthesis straight x squared minus straight y squared right parenthesis plus 2 straight y left parenthesis straight x squared minus straight y squared right parenthesis dy over dx therefore space space space space space open square brackets 2 space straight y space left parenthesis straight x squared minus straight y squared right parenthesis space plus space straight y left parenthesis straight x squared plus straight y squared right parenthesis close square brackets dy over dx space equals space straight x left parenthesis straight x squared plus straight y squared right parenthesis minus 2 straight x left parenthesis straight x squared minus straight y squared right parenthesis therefore space space space left parenthesis 2 straight x squared straight y minus 2 straight y cubed plus straight x squared straight y plus straight y cubed right parenthesis dy over dx space equals space straight x cubed plus xy squared minus 2 straight x cubed plus 2 xy squared therefore space space space space left parenthesis 3 straight x squared straight y minus straight y cubed right parenthesis dy over dx space equals space 3 xy squared minus straight x cubed therefore space space space space space space space space space space space space space space space space space space space space space space space space space space space dy over dx space equals space fraction numerator straight x cubed minus 3 xy squared over denominator straight y cubed minus 3 straight x squared straight y end fraction space which space is space the space required space equation.$
Hence the result.