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Differential Equations

Short Answer Type

221.

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

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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:
$straight x squared dy plus left parenthesis xy plus straight y squared right parenthesis dx space equals space 0 semicolon space space space straight y space equals space 1 space when space straight x space equals space 1$

The given differential equation is

straight x squared dy space plus straight y left parenthesis straight x plus straight y right parenthesis space dx space equals space 0 comma space space space or space space space straight x squared space dy space equals space minus straight y left parenthesis straight x plus straight y right parenthesis space dx

therefore space space space space space dy over dx space equals space minus fraction numerator straight y left parenthesis straight x plus straight y right parenthesis over denominator straight x squared end fraction

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

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

therefore space space space space from space left parenthesis 1 right parenthesis comma space space space straight v plus straight x dv over dx space equals space minus fraction numerator straight v space straight x left parenthesis straight x plus vx right parenthesis over denominator straight x squared end fraction space space space space space rightwards double arrow space space space space straight v plus straight x dv over dx equals negative straight v minus straight v squared

rightwards double arrow space space space space straight x space dv over dx space equals space straight v squared minus 2 straight v

therefore space space space space space fraction numerator 1 over denominator straight v squared plus 2 straight v end fraction dv space equals space minus 1 over straight x dx space space space space space rightwards double arrow space space space integral fraction numerator 1 over denominator straight v left parenthesis straight v plus 2 right parenthesis end fraction dv space equals space minus integral 1 over straight x dx rightwards double arrow space space space space integral open square brackets fraction numerator 1 over denominator straight v left parenthesis 0 plus 2 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis negative 2 right parenthesis thin space left parenthesis straight v plus 2 right parenthesis end fraction close square brackets dv space equals space minus integral 1 over straight x dx rightwards double arrow space space space space 1 half integral open parentheses 1 over straight v minus fraction numerator 1 over denominator straight v plus 2 end fraction close parentheses space dv space equals space minus integral 1 over straight x dx therefore space space space space space 1 half open square brackets log space straight v space minus space log space left parenthesis straight v plus 2 right parenthesis close square brackets space equals space minus log space straight x space plus space log space straight c rightwards double arrow space space space space space 1 half log space open parentheses fraction numerator straight v over denominator straight v plus 2 end fraction close parentheses space equals space log space open parentheses straight c over straight x close parentheses therefore space space space space space fraction numerator begin display style straight y over straight x end style over denominator begin display style straight y over straight x end style plus 2 end fraction space equals space straight c squared over straight x squared space space space space rightwards double arrow space space space space space fraction numerator straight y over denominator 2 straight x plus straight y end fraction space equals space straight c squared over straight x squared rightwards double arrow space space space space space space space straight x squared straight y space equals space straight k space left parenthesis 2 straight x plus straight y right parenthesis comma space where space straight k space equals space straight c squared 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 2 right parenthesis space space space space space space space space space space space space space space space space space Now space straight y space equals space 1 space when space straight x space equals space 1 space space space space space space rightwards double arrow space space space space space 1 space equals space straight k left parenthesis 2 plus 1 right parenthesis space space space rightwards double arrow space space space straight k space equals space 1 third

therefore space space space space from space left parenthesis 2 right parenthesis comma space space straight x squared straight y space equals space 1 third left parenthesis 2 straight x plus straight y right parenthesis space which space is space required space solution. space

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:
$dy over dx minus straight y over straight x plus cosec space open parentheses straight y over straight x close parentheses space equals space 0 space semicolon space space space straight y space equals space 0 space space space when space straight x space equals space 1$

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:
$2 xy plus straight y squared minus 2 straight x squared straight y apostrophe space equals space 0 space space space space space space space space space space space space space straight y left parenthesis 1 right parenthesis space equals space 2$