Find a one parameter family of solutions of each of the followin
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    Differential Equations

    Multiple Choice QuestionsLong Answer Type

    281. Find the general solution of the following differential equation:
    dy over dx plus fraction numerator 4 straight x over denominator straight x squared plus 1 end fraction straight y space equals negative fraction numerator 1 over denominator left parenthesis straight x squared plus 1 right parenthesis cubed end fraction
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    Multiple Choice QuestionsShort Answer Type

    282.

    Solve:  open parentheses straight x plus straight y close parentheses space dy over dx space equals space 1.

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    Answer

    Multiple Choice QuestionsLong Answer Type

    283.

    Solve:
          open parentheses straight x plus 2 straight y cubed close parentheses space straight y apostrophe space equals space straight y comma space space straight y greater than 0

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    Answer

    Multiple Choice QuestionsShort Answer Type

    284.

    Solve:
    straight y space dx space plus space left parenthesis straight x minus straight y right parenthesis squared space dy space equals space 0
          

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    Multiple Choice QuestionsLong Answer Type

    285. Find the general solution of the differential equation y dx – (x + 2 y2) dy = 0.
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    286. Solve the following differential equation:
    (1 + y2)dx = (tan– 1 y – x) dy
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    287. Solve the following differential equation:
    square root of 1 minus straight y squared end root space dx equals left parenthesis sin to the power of negative 1 end exponent straight y space minus straight x right parenthesis space dy
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    Answer

    Multiple Choice QuestionsShort Answer Type

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    288. Find a one parameter family of solutions of each of the following differential equation e–y sec2 y dy = dx + x dy.

    The given differential equation is
                       e–y sec2 y dy = dx + x dy
    therefore space space space left parenthesis straight e to the power of negative straight y end exponent space sec squared straight y minus straight x right parenthesis space dy space equals space dx space space space space space space space space space space space or space space space space dx over dy space equals space straight e to the power of negative straight y end exponent space sec squared straight y minus straight x therefore space space space space space space space space dy over dx plus straight x space equals space straight e to the power of negative straight y end exponent space secy
    Comparing it with dx over dy plus Px space equals space straight Q comma space space we space get space straight P space equals space 1 comma space space straight Q space equals space straight e to the power of negative straight y end exponent space sec squared straight y
                  integral straight P space dy space equals space integral 1 space dy space equals space straight y. space straight e to the power of integral straight P space dy end exponent space equals space straight e to the power of straight y
    therefore    solution of differential equation is
                                        xe to the power of integral straight P space dy end exponent space equals space integral Qe to the power of integral straight P space dy end exponent plus straight c space space space space space space or space space space space space straight x space straight e to the power of straight y space equals space integral straight e to the power of negative straight y end exponent space sec squared straight y. space straight e to the power of straight y space dy space plus straight c
    or                               straight x space straight e to the power of straight y space equals space integral sec squared straight y space dy space plus space straight c space space space space space space or space space space straight x space straight e to the power of straight y space equals space tan space straight y space plus straight c

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    Multiple Choice QuestionsLong Answer Type

    289. Find the particular solution of the differential equation:
                         dy over dx plus straight y space cotx space equals space 2 straight x plus straight x squared space cotx space space space left parenthesis straight x not equal to 0 right parenthesis
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    290. For the given differential equation, find a particular solution satisfying the given condition:
    dy over dx plus 2 straight y space tan space straight x space equals space sin space straight x space colon space straight y space space equals space 0 space space when space straight x space equals space straight pi over 3

                         
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    Answer
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