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Differential Equations

Question

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Show that the differential equation $straight x space cos space open parentheses straight y over straight x close parentheses dy over dx space equals straight y space cos space open parentheses straight y over straight x close parentheses plus straight x$ is homogeneous and solve it.

Solution

The given differential equation is

space straight x space cos space open parentheses straight y over straight x close parentheses space dy over dx space equals space straight y space cos space open parentheses straight y over straight x close parentheses plus space straight x space space space or space space space space dy over dx space equals fraction numerator straight y space cos space open parentheses begin display style straight y over straight x end style close parentheses plus straight x over denominator straight x space cos space open parentheses begin display style straight y over straight x end style close parentheses end fraction

...(1)
It is 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 y space cos space open parentheses begin display style straight y over straight x end style close parentheses plus straight x over denominator straight x space cos space open parentheses begin display style straight y over straight x end style close parentheses end fraction

Replacing x by

straight lambda space straight x

and y by

λy comma

we get

space straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator straight lambda space open square brackets straight y space cos space open parentheses begin display style straight y over straight x end style close parentheses plus straight x close square brackets over denominator straight lambda space open parentheses straight x space cos space begin display style straight y over straight x end style close parentheses end fraction space equals space straight lambda degree space 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

the given differential equation is a homogeneous differential equation
Put y = vx that

dy over dx equals 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 v space straight x space cos space open parentheses begin display style vx over straight x end style close parentheses plus straight x over denominator straight x space cos space open parentheses begin display style vx over straight x end style close parentheses end fraction

therefore space space space straight v plus straight x dv over dx space equals space fraction numerator straight v space straight x space cos space straight v space plus space straight x over denominator straight x space cos space straight v end fraction

or

straight v plus straight x dv over dx space equals fraction numerator straight v space cos space straight v plus 1 over denominator cos space straight v end fraction space space or space space straight x dv over dx space equals space fraction numerator straight v space cos space straight v space plus space 1 over denominator cos space straight v end fraction minus straight v

therefore space space space space space space space space straight x dv over dx space equals space fraction numerator straight v space cos space straight v plus 1 minus straight v space cos space straight v over denominator cos space straight v end fraction therefore space space space space straight x dv over dx equals fraction numerator 1 over denominator cos space straight v end fraction

Separating the variables, we get,

cosv space dv space equals space 1 over straight x dx

Integrating,

integral space cos space straight v space dv space equals space integral 1 over straight x dx

therefore

sin space straight v space equals space log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight c close vertical bar

or

sin space straight y over straight x space equals space log space open vertical bar straight c space straight x close vertical bar space which space is space required space solution. space

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