Differential Equations

Question

For the given differential equation, find the particular solution satisfying the given condition:
$open square brackets straight x space sin squared open parentheses straight y over straight x close parentheses minus straight y close square brackets space dx plus space straight x space dy space equals space 0 colon space space space straight y space equals straight pi over 4 space space when space straight x space equals space 1$

Solution

The given differential equation is

open square brackets straight x space sin squared space open parentheses straight y over straight x close parentheses minus straight y close square brackets space dx plus straight x space dy space equals space 0

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

dy over dx space equals straight y over straight x minus sin squared space open parentheses straight y over straight x close parentheses

...(1)
Put y = v x so that

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

therefore space space space space space space left parenthesis 1 right parenthesis space becomes comma space space space straight v plus straight x dv over dx equals straight v minus sin squared straight v or space space space space space space space space space straight x dv over dx equals negative sin squared straight v

Separating the variables,

fraction numerator 1 over denominator sin squared straight v end fraction dv space equals space minus dx over straight x

Integrating,

integral cosec squared straight v space dv space equals space minus integral dx over straight x

rightwards double arrow space space space log space open vertical bar straight x close vertical bar minus cot space straight v space equals space straight c rightwards double arrow space space space log space open vertical bar straight x close vertical bar minus cot space open parentheses straight y over straight x close parentheses space equals space straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

Now, x = 1, when

straight y space equals space straight pi over 4

log space open vertical bar 1 close vertical bar space minus space cot space open parentheses straight pi over 4 close parentheses space equals straight c

rightwards double arrow space space space space straight c space equals space 0 minus 1 space equals space minus 1

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log space open vertical bar straight x close vertical bar space minus space cot space straight y over straight x space equals space space minus space log space straight e

log space open vertical bar straight e space straight x close vertical bar space equals space cot space open parentheses straight y over straight x close parentheses

.
which is the required particular solution.

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