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Differential Equations

Short Answer Type

221.

Solve:
$straight y apostrophe space equals space straight y over straight x plus sin straight y over straight x$

78 Views

222.

Solve:
$dy over dx space equals straight y over straight x plus tan straight y over straight x open parentheses 0 less than straight y over straight x less than straight pi over 2 close parentheses$

74 Views

Long Answer Type

223.

For the given differential equations, find the particular solution satisfying the given condition:
$left parenthesis straight x plus straight y right parenthesis space dy space plus space left parenthesis straight x minus straight y right parenthesis dx space equals space 0 space semicolon space space straight y space equals space 1 space space space when space straight x space equals space 1$

75 Views

224.

For the given differential equation, find the particular solution satisfying the given condition:
$straight x squared dy plus left parenthesis xy plus straight y squared right parenthesis dx space equals space 0 semicolon space space space straight y space equals space 1 space when space straight x space equals space 1$

80 Views

225.

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

73 Views

226.
For the given differential equation, find the particular solution satisfying the given condition:
$dy over dx minus straight y over straight x plus cosec space open parentheses straight y over straight x close parentheses space equals space 0 space semicolon space space space straight y space equals space 0 space space space when space straight x space equals space 1$

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

75 Views

227.

For the given differential equation, find the particular solution satisfying the given condition:
$2 xy plus straight y squared minus 2 straight x squared dy over dx equals 0 semicolon space space space straight y space equals 2 space space when space straight x space equals space 1.$

75 Views

228.

For the given differential equation, find the particular solution satisfying the given condition:
$2 straight x squared straight y apostrophe space minus space 2 xy plus straight y squared space equals space 0 comma space space space space straight y left parenthesis straight e right parenthesis space equals space straight e$

74 Views

229.

For the given differential equation, find the particular solution satisfying the given condition:
$2 xy plus straight y squared minus 2 straight x squared straight y apostrophe space equals space 0 space space space space space space space space space space space space space straight y left parenthesis 1 right parenthesis space equals space 2$