Differential Equations

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Question

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
$1 over straight x cos straight y over straight x dx minus open parentheses straight x over straight y sin straight y over straight x plus cos straight y over straight x close parentheses dy space equals 0$

Solution

The given differential equation is

straight y over straight x cos straight y over straight x dx minus open parentheses straight x over straight y sin straight y over straight x plus cos straight y over straight x close parentheses dy space equals space 0

dy over dx space equals space fraction numerator begin display style straight y over straight x end style cos begin display style straight y over straight x end style over denominator begin display style straight x over straight y end style sin begin display style straight y over straight x end style plus cos begin display style straight y over straight x end style end fraction

Put

straight y space equals space space straight v space straight x space so space that space dy over dx space equals space straight v plus straight x dv over dx

therefore space space space space space space space space space space straight v plus straight x dv over dx space equals space fraction numerator straight v space cos space straight v over denominator begin display style 1 over straight v end style sinv space plus cos space straight v end fraction space space space rightwards double arrow space space space space straight v plus straight x dv over dx space equals fraction numerator straight v squared space cosv over denominator sin space straight v plus straight v space cos space straight v end fraction therefore space space space space space space space straight x dv over dx space equals space fraction numerator straight v squared space cos space straight v over denominator sin space straight v plus straight v space cos space straight v end fraction minus straight v therefore space space space space space straight x dv over dx space equals space fraction numerator straight v squared space cos space straight v minus space straight v space sin space straight v space minus space straight v squared space cos space straight v over denominator sin space straight v plus straight v space cos space straight v end fraction therefore space space space space straight x dv over dx space equals negative fraction numerator straight v space sin space straight v over denominator sin space straight v plus straight v space cos space straight v end fraction

Separating the variables and integrating ,we get,

integral fraction numerator sin space straight v plus space straight v space cosv over denominator straight v space sinv end fraction dv space equals negative integral 1 over straight x dx

therefore space space log space open vertical bar straight v space sinv close vertical bar space equals space minus space log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight A close vertical bar rightwards double arrow space space space log space open vertical bar straight y over straight x sin space straight y over straight x close vertical bar plus space log space open vertical bar straight x close vertical bar space equals space log space open vertical bar straight A close vertical bar rightwards double arrow space space log space open vertical bar straight y space sin straight y over straight x close vertical bar space minus space log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight x close vertical bar space equals space log space open vertical bar straight A close vertical bar rightwards double arrow space space log space open vertical bar straight y space sin straight y over straight x close vertical bar space equals space log space open vertical bar straight A close vertical bar rightwards double arrow space space space space space space straight y space sin space straight y over straight x space equals space straight A comma space space space which space is space required space solution. space