Solve:       from Mathematics Differential Equations
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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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    Answer

    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

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    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

    The given equation is
                  left parenthesis straight x plus 2 straight y cubed right parenthesis space dy over dx space equals space straight y

              or space space open parentheses straight x plus 2 straight y cubed close parentheses. space space fraction numerator 1 over denominator begin display style dx over dy end style end fraction space equals space straight y space space space space space space or space space space space straight x plus 2 straight y cubed space equals space straight y space dy over dx or space space straight x over straight y plus 2 straight y squared space equals space dx over dy space space space space space space space or space space space space space space space dy over dx minus 1 over straight y. space straight x space equals space 2 space straight y squared
    Comparing it with dx over dy plus Px space equals space straight Q comma space space we space get comma
                 straight P space equals negative 1 over straight y comma space space straight Q space equals space 2 straight y squared integral straight P space dy space equals space minus integral 1 over straight y dy space equals space minus log space straight y
    straight I. straight F. space equals space straight e to the power of integral space straight P space dy end exponent space equals space straight e to the power of negative log space straight y end exponent space equals space straight e to the power of log space straight y to the power of negative 1 end exponent end exponent space equals space straight y to the power of negative 1 end exponent space equals space 1 over straight y
    therefore   solution of given equation is
                      straight x. space 1 over straight y space equals space integral space 2 straight y squared. space 1 over straight y dy 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 open square brackets because space space straight x. space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent dy plus straight c close square brackets

    or space space space space straight x over straight y space equals space 2 integral straight y space dy space plus straight c or space space space straight x over straight y space equals space straight y squared plus straight c comma space space or space space straight x space equals space straight y cubed plus straight c space straight y
    which is required solution. 

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    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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    Multiple Choice QuestionsShort Answer Type

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