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Differential Equations

Question
CBSEENMA12033150

Show that the family of curves for which the slope of the tangent at any point (x, y) on it is fraction numerator straight x squared plus straight y squared over denominator 2 xy end fraction comma is given by straight x squared minus straight y squared space equals space straight c space straight x.

Solution

We know that dy over dx is slope of tangent to the curve at point (x, y).
therefore space space space space space space space space space dy over dx space equals space fraction numerator straight x squared plus straight y squared over denominator 2 xy end fraction
  Put y = v x so that dy over dx equals straight v plus straight x dv over dx
therefore space space space space straight v plus straight x dv over dx equals fraction numerator straight x squared plus straight v squared straight x squared over denominator 2 vx squared end fraction
therefore space space space space straight v plus straight x dv over dx space equals fraction numerator 1 plus straight v squared over denominator 2 straight v end fraction
therefore space space space space space space space space space straight x dv over dx space equals space fraction numerator 1 plus straight v squared over denominator 2 straight v end fraction minus straight v space space space space space space or space space space space space straight x dv over dx equals space fraction numerator 1 minus straight v squared over denominator 2 straight v end fraction
Separating the variables and integrating, we get,
             integral fraction numerator 2 straight v over denominator 1 minus straight v squared end fraction dv space equals space integral 1 over straight x dx space space space space or space space space integral fraction numerator 2 straight v over denominator straight v squared minus 1 end fraction dv space equals space minus integral 1 over straight x dx
therefore space space space log space open vertical bar straight v squared minus 1 close vertical bar space equals space minus log space open vertical bar straight x close vertical bar plus log space open vertical bar straight c subscript 1 close vertical bar space space space or space space space log space open vertical bar left parenthesis straight v squared minus 1 right parenthesis thin space left parenthesis straight x right parenthesis close vertical bar space equals space log space open vertical bar straight c subscript 1 close vertical bar
or space space space space space space space left parenthesis straight v squared minus 1 right parenthesis space straight x space equals space plus-or-minus space space straight c subscript 1
Replacing v by  straight y over straight x comma space we get
                        open parentheses straight y squared over straight x squared minus 1 close parentheses space straight x space equals space plus-or-minus space straight c subscript 1 space space space or space space space space left parenthesis straight y squared minus straight x squared right parenthesis space equals space space plus-or-minus space straight c subscript 1 straight x space space space or space space space straight x squared minus straight y squared space equals space cx.

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