Find one-parameter families of solution curves of the following

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 Multiple Choice QuestionsLong Answer Type

251.

Solve:
straight x space log space straight x space dy over dx plus straight y space equals space 2 over straight x log space straight x

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 Multiple Choice QuestionsShort Answer Type

252.

Find one-parameter families of solution curves of the following differential equations:
straight y apostrophe plus 3 straight y space equals straight e to the power of negative 2 straight x end exponent

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253.

Find one-parameter families of solution curves of the following differential equation:
 straight y apostrophe space minus space straight y space equals space cos space straight x

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254.

Find one-parameter families of solution curves of the following differential equation:
 y'+3 y = emx (m: a given real number)

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255.

Find one-parameter families of solution curves of the following differential equation:
 y' - y = cos 2x

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256.

Find one-parameter families of solution curves of the following differential equation:
x y' - y = (x+1) e-x

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257.

Find one-parameter families of solution curves of the following differential equation:
x y' + y = x4

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258.

Find one-parameter families of solution curves of the following differential equation:
x y' log x + y = log x

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259.

Find one-parameter families of solution curves of the following differential equation:
dy over dx minus fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction comma space space straight y space equals space straight x squared plus 2


The given differential equation is
                dy over dx minus fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction straight y space equals space straight x squared plus 2
Comparing it with dy over dx plus Py space equals space straight Q comma space we space get comma space straight P space equals negative fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction comma space space straight Q space equals straight x squared plus 2
therefore space space space space space integral straight P space dx space equals space minus integral fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction dx equals negative log space left parenthesis 1 plus straight x squared right parenthesis
therefore space space space space space straight e to the power of integral space Pdx end exponent space equals space straight e to the power of negative log space left parenthesis 1 plus straight x squared right parenthesis end exponent space equals space straight e to the power of log space left parenthesis 1 plus straight x squared right parenthesis to the power of negative 1 end exponent end exponent space equals space fraction numerator 1 over denominator 1 plus straight x squared end fraction
therefore  solution of differential equation is                                straight y space straight e to the power of integral straight P space dx end exponent space equals integral straight Q. space straight e to the power of integral straight P space dx end exponent dx space plus space straight c space space space or space space straight y. space fraction numerator 1 over denominator 1 plus straight x squared end fraction space equals space integral left parenthesis straight x squared plus 2 right parenthesis. space fraction numerator 1 over denominator straight x squared plus 1 end fraction dx plus straight c
or       fraction numerator straight y over denominator 1 plus straight x squared end fraction space equals integral fraction numerator straight x squared plus 2 over denominator straight x squared plus 1 end fraction dx plus straight c    or    fraction numerator straight y over denominator 1 plus straight x squared end fraction space equals space integral open parentheses 1 plus fraction numerator 1 over denominator straight x squared plus 1 end fraction close parentheses dx plus straight c
or              fraction numerator straight y over denominator 1 plus straight x squared end fraction space equals space straight x plus tan to the power of negative 1 end exponent straight x plus straight c
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 Multiple Choice QuestionsLong Answer Type

260.

Find one-parameter families of solution curves of the following differential equation:
  straight y apostrophe plus straight y space cosx space equals space straight e to the power of sin space straight x end exponent space cos space straight x

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