Find a particular solution of the differential equation(x – y)

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 Multiple Choice QuestionsShort 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

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 Multiple Choice QuestionsLong 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

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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

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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


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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




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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.





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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
       





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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
          
       





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 Multiple Choice QuestionsShort Answer Type

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230.

Find a particular solution of the differential equation
(x – y) (dx + dy) = dx – dy. given that y = – 1, when x = 0.


The given differential equation is
              (x – y) (dx + dy) = dx – dy                          ...(1)
or             dx plus dy space equals space fraction numerator dx minus dy over denominator straight x minus straight y end fraction
Integrating, we get
                      integral left parenthesis dx plus dy right parenthesis space equals space integral fraction numerator straight d space left parenthesis straight x minus straight y right parenthesis over denominator straight x minus straight y end fraction plus straight c
rightwards double arrow space space space space space space space space space space space space straight x plus straight y space equals space log space open vertical bar straight x minus straight y close vertical bar plus straight c                        ....(2)
Now x = 0,  y = -1
therefore space space space space space space space 0 plus left parenthesis negative 1 right parenthesis space equals space log space open vertical bar 0 plus 1 close vertical bar space plus space straight c
rightwards double arrow space space space space space space space space space space space straight c space equals space minus 1
therefore space space space from space left parenthesis 2 right parenthesis comma space space space straight x plus straight y space equals space log space open vertical bar straight x minus straight y close vertical bar space minus space 1
which is required solution.

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