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Differential Equations

Question
CBSEENMA12033143
Wired Faculty App

For the given differential equation, find the particular solution satisfying the given condition:
dy over dx minus straight y over straight x plus cosec space open parentheses straight y over straight x close parentheses space equals space 0 space semicolon space space space straight y space equals space 0 space space space when space straight x space equals space 1




Solution
The given differential equation is
                                             dy over dx space equals space straight y over straight x minus cosec space straight y over straight x
Put y = v x  so that dy over dx space equals space straight v plus straight x dv over dx
therefore space space space space space space space space straight v plus straight x dv over dx space equals space straight v minus cosec space straight v therefore space space space space space space space space space space space space space straight x dv over dx space equals space minus cosec space straight v comma space space space space space space space space space space space space space space space space therefore space space space space minus fraction numerator 1 over denominator cosec space straight v end fraction dv space equals space 1 over straight x dx therefore space space space space space minus integral sin space straight v space dv space equals space integral 1 over straight x dx therefore space space space space space space space cos space straight v space space equals space log space open vertical bar straight x close vertical bar plus straight c therefore space space space space space cos space straight y over straight x space equals space log space open vertical bar straight x close vertical bar plus 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 space space space space space space space space space... left parenthesis 1 right parenthesis space space space space Now space straight y space equals space 0 space space space space when space straight x space space equals space 1 therefore space space space space space space space cos space 0 space equals space log space open vertical bar 1 close vertical bar space plus space straight c space space space space space space rightwards double arrow space space space space space 1 space equals space 0 plus space straight c space rightwards double arrow space space space straight c space space equals 1 therefore space space from space left parenthesis 1 right parenthesis comma space space cos space straight y over straight x space equals space log space open vertical bar straight x close vertical bar plus 1 space is space the space required space solution. space

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