For the following differential equation, find the general soluti
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    Differential Equations

    Multiple Choice QuestionsShort Answer Type

    121. Find a one parameter family of solutions of each of the following differential equation:
     (x y2 + 2x) dx + (x2 y + 2y) dy = 0
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    Answer
    122. Find a one parameter family of solutions of each of the following differential equation:
    x y y' = 1 + x + y + x y
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    Answer
    123.

    Solve:
    dy over dx equals straight e to the power of straight x minus straight y end exponent plus straight x squared straight e to the power of straight y

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    Answer
    124.

    Solve:
    dy over dx space equals straight e to the power of straight x plus straight y end exponent plus straight x squared space straight e to the power of straight y.

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    Answer
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    125.

    Solve:
    dy over dx space equals straight e to the power of straight x minus straight y end exponent plus straight x cubed straight e to the power of negative straight y end exponent

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    Answer
    126.

    Solve:
    x (1 + y2) dx + y (1 + x2)dy = 0  

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    Answer
    127.

    Solve:
    (1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0.

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    Answer
    128.

    For the following differential equation, find the general solution:
    dy over dx equals fraction numerator 1 minus cosx over denominator 1 plus cosx end fraction

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    129.

    For the following differential equation, find the general solution:
    dy over dx space equals sin to the power of negative 1 end exponent straight x

    The given differential equation is
                        dy over dx space equals space sin to the power of negative 1 end exponent straight x
    Separating the variables and integrating,
                           integral space dy space equals space integral space sin to the power of negative 1 end exponent straight x space dx
    therefore space space space space space space space space space space space space integral space 1 space dy space equals space integral space sin to the power of negative 1 end exponent straight x. space 1 space dx therefore space space space space space space space space space space straight y space equals space left parenthesis sin to the power of negative 1 end exponent straight x right parenthesis. space straight x space minus space space integral fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction. straight x space dx or space space space space space space space space space space space space space space space space space space space straight y space equals space straight x space sin to the power of negative 1 end exponent straight x space plus space 1 half integral left parenthesis 1 minus straight x squared right parenthesis to the power of negative 1 half end exponent left parenthesis negative 2 space straight x right parenthesis space dx therefore space space space space space space space space space space space space space space space space space space space straight y space equals space straight x space sin to the power of negative 1 end exponent straight x space plus space 1 half fraction numerator left parenthesis 1 minus straight x squared right parenthesis to the power of begin display style 1 half end style end exponent over denominator begin display style 1 half end style end fraction plus straight c therefore space space space space space space space space space space space space space space space space space space space space straight y space equals space straight x space sin to the power of negative 1 end exponent straight x space plus space square root of 1 minus straight x squared end root plus straight c
    which is the required solution.
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    130. Solve 3ex tan y dx + (1 + ex) sec2 y dy = 0. 
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    Answer
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