Differential Equations

Question

Show that the given differential equation is homogeneous and solve it:
$open parentheses straight x squared minus straight y squared close parentheses space dx space plus space 2 xy space dy space equals space 0$

Solution

The given differential equation is
$open parentheses straight x squared minus straight y squared close parentheses dx plus 2 xy space dy space equals space 0 space space or space space space 2 xy space dy space equals space left parenthesis straight y squared minus straight x squared right parenthesis space dx$
or $dy over dx space equals fraction numerator straight y squared minus straight x squared over denominator 2 xy end fraction$
It is a differential equation of the form $dy over dx space equals straight F left parenthesis straight x comma space straight y right parenthesis$
Here, $straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator straight y squared minus straight x squared over denominator 2 space straight x space straight y end fraction$
Replacing x by $λx$ and y by $λy comma$ we get,
$straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator straight lambda squared straight y squared minus straight lambda squared straight x squared over denominator 2 space straight lambda squared space straight x space straight y end fraction space equals space fraction numerator straight lambda squared space left parenthesis straight y squared minus straight x squared right parenthesis over denominator straight lambda squared space left parenthesis 2 xy right parenthesis end fraction space equals space straight lambda degree space space left square bracket straight F left parenthesis straight x comma space straight y right parenthesis right square bracket$
∴ F(x, y) is a homogeneous function of degree zero.
∴ given differential equation is a homogeneous differential equation.
Put y = vx so that $dy over dx equals straight v plus straight x dv over dx$
$therefore space space space space space straight v plus straight x dv over dx equals fraction numerator straight v squared straight x squared minus straight x squared over denominator 2 space straight v space straight x squared end fraction space space or space space space straight v plus straight x dv over dx space equals space fraction numerator straight v squared minus 1 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 straight v squared minus 1 over denominator 2 straight v end fraction minus straight v space space space or space space space straight x dv over dx space equals fraction numerator straight v squared minus 1 minus 2 straight v squared over denominator 2 straight v end fraction therefore space space space space space space straight x dv over dx space equals fraction numerator negative 1 minus straight v squared over denominator 2 straight v end fraction space space space space rightwards double arrow space space space space space space fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space minus 1 over straight x space dx therefore space space space space space integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space minus integral 1 over straight x dx therefore space space space space space log space open vertical bar 1 plus straight v squared close vertical bar space equals space minus log space open vertical bar straight x close vertical bar plus straight c apostrophe therefore space space space space log space open vertical bar 1 plus straight v squared close vertical bar space plus space log space open vertical bar straight x close vertical bar space equals space space straight c apostrophe therefore space space log space open vertical bar left parenthesis 1 plus straight v squared right parenthesis thin space left parenthesis straight x right parenthesis close vertical bar space equals space straight c apostrophe therefore space space space space space space space space straight x left parenthesis 1 plus straight v squared right parenthesis space equals space straight e to the power of straight c apostrophe end exponent space space space space space space space space rightwards double arrow space space space space space space space space straight x open parentheses 1 plus straight y squared over straight x squared close parentheses space equals space straight c therefore space space space space straight x squared plus straight y squared space equals space straight c space straight x space$
is the required solution.

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Differential Equations

Show that the given differential equation is homogeneous and solve it:
$open parentheses straight x squared minus straight y squared close parentheses space dx space plus space 2 xy space dy space equals space 0$

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