Differential Equations

Question

Show that the given differential equation is homogeneous and solve it:
$left parenthesis straight x squared plus xy right parenthesis space dy space equals space left parenthesis straight x squared plus straight y squared right parenthesis space dx$

Solution

The given differential equation is

open parentheses straight x squared plus straight x space straight y close parentheses dy space equals space left parenthesis straight x squared plus straight y squared right parenthesis space dx

dy over dx space equals space fraction numerator straight x squared plus straight y squared over denominator straight x squared plus xy end fraction

...(1)
It is a differential equation of the form

dy over dx space equals space 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 x squared plus straight y squared over denominator straight x squared plus straight x space straight y end fraction

Replacing x by

λx space and space straight y space by space λy comma

we get

straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator straight lambda squared straight x squared plus straight lambda squared straight y squared over denominator straight lambda squared straight x squared plus straight lambda squared xy end fraction space equals space fraction numerator straight lambda squared left parenthesis straight x squared plus straight y squared right parenthesis over denominator straight lambda squared left parenthesis straight x squared plus xy right parenthesis end fraction space equals space straight lambda degree space left square bracket straight F space left parenthesis straight x comma space straight y right parenthesis right square bracket

therefore space space space straight F left parenthesis straight x comma space straight y right parenthesis

is a homogeneous function of degree zero.

therefore space space space

given differential equation is a homogeneous differential equation.
Put y = v x so that

dy over dx space equals space straight v plus straight x dv over dx

therefore space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx space equals space fraction numerator straight x squared plus straight v squared straight x squared over denominator straight x squared plus vx squared end fraction rightwards double arrow space space space straight v plus straight x dv over dx space equals space fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction space space space space rightwards double arrow space space space straight x dv over straight d space equals space fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction minus straight v therefore space space space space straight x dv over dx space equals space fraction numerator 1 minus straight v over denominator 1 plus straight v end fraction space space space space rightwards double arrow space space space fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction dv space equals space 1 over straight x dx space space space rightwards double arrow space space space space integral open parentheses negative 1 plus fraction numerator 2 over denominator 1 minus straight v end fraction close parentheses dv space equals space integral 1 over straight x dx therefore space space space space space minus straight v minus 2 space log space left parenthesis 1 minus straight v right parenthesis space equals space log space straight x space plus straight c rightwards double arrow space space space minus straight y over straight x minus 2 log space open parentheses 1 minus straight y over straight x close parentheses space equals space log space straight x space plus space straight c rightwards double arrow space space space space minus straight y minus 2 straight x space log space open parentheses fraction numerator straight x minus straight y over denominator straight x end fraction close parentheses space equals space straight x space log space straight x space plus space straight c space straight x

which is required solution.

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Differential Equations

Show that the given differential equation is homogeneous and solve it:
$left parenthesis straight x squared plus xy right parenthesis space dy space equals space left parenthesis straight x squared plus straight y squared right parenthesis space dx$

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