Question

Find the particular solution of (1 + x² + y² + x² y² ) dx + x y dy = 0

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Solution

The given differential equation is
(1 + x² + y² + x² y² ) dx + x y dy = 0
or

open square brackets left parenthesis 1 plus straight x squared right parenthesis space plus space straight y squared left parenthesis 1 plus straight x squared right parenthesis close square brackets dx space plus space straight x space straight y space dy space equals space 0

left parenthesis 1 plus straight x squared right parenthesis space left parenthesis 1 plus straight y squared right parenthesis space dx space space plus straight x space straight y space dy space equals space 0

straight x space straight y space dy space equals space minus left parenthesis 1 plus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis space dx

therefore space space space space space fraction numerator straight y over denominator 1 plus straight y squared end fraction dy space equals space minus fraction numerator 1 plus straight x squared over denominator straight x end fraction dx therefore space space space space fraction numerator straight y over denominator 1 plus straight y squared end fraction dy space equals space minus open parentheses 1 over straight x plus straight x close parentheses dx

Integrating,

1 half integral fraction numerator 2 straight y over denominator 1 plus straight y squared end fraction dy space equals space minus integral open parentheses 1 over straight x plus straight x close parentheses space dx

therefore space space space space 1 half space log space left parenthesis 1 plus straight y squared right parenthesis space equals space minus open square brackets log space open vertical bar straight x close vertical bar plus straight x squared over 2 close square brackets plus straight c

...(1)
Now

straight y space equals space 0 space space when space straight x space equals space 1

therefore space space space space space space space space space space space space space space space 1 half log space 1 space space equals space minus open square brackets log space 1 plus 1 half close square brackets plus straight c therefore space space space space space space space space space space space space space space space space 0 space equals space minus 1 half plus straight c space space space space space space space space space space space space space space rightwards double arrow space space space straight c space equals space 1 half therefore space space space space from space left parenthesis 1 right parenthesis space space space space space space space space space space space space space 1 half log space open square brackets 1 plus straight y squared close square brackets equals space minus open square brackets log space open vertical bar straight x close vertical bar plus straight x squared over 2 close square brackets plus 1 half

which is required solution.

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