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111. Solve the following differential equation:

open parentheses straight x squared straight y plus straight x squared close parentheses space dx space plus space left parenthesis straight x space straight y squared minus straight y squared right parenthesis space dy space equals space 0

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112. Solve the following differential equation:

straight y left parenthesis 1 minus straight x squared right parenthesis dy over dx space equals space straight x left parenthesis 1 plus straight y squared right parenthesis

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113. Find the general solution of each of the following equations

dy over dx equals straight x to the power of 5 plus straight x squared minus 2 over straight x

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114. Find the general solution of each of the following equations:

left parenthesis straight x plus 2 right parenthesis space dy over dx space equals space straight x squared plus 4 straight x minus 9

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115. Find the general solution of each of the following equations:

dy over dx plus 2 straight x equals space straight e to the power of 3 straight x end exponent

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116. Find the general solution of each of the following equations:

dy over dx equals fraction numerator straight y left parenthesis straight x squared minus 1 right parenthesis over denominator straight x left parenthesis straight y squared minus 1 right parenthesis end fraction

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117. Find the general solution of each of the following equation:

left parenthesis straight x minus 1 right parenthesis space straight y apostrophe space equals space 2 space straight x cubed space straight y space where space straight y apostrophe space equals space dy over dx

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118. Find the general solution of the differential equation:

dy over dx plus square root of fraction numerator 1 minus straight y squared over denominator 1 minus straight x squared end fraction end root space equals space 0.

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119. Find a one parameter family of solutions of each of the following differential equation:
y' = e^{x + y} + e^{–x + y}

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The given differential equation is

dy over dx space equals space left parenthesis cos squared straight x minus sin squared straight x right parenthesis space cos squared straight y

Separating the variables, we get,

fraction numerator 1 over denominator cos squared straight y end fraction dy space equals space left parenthesis cos squared straight x minus sin squared straight x right parenthesis space dx

Integrating,

integral sec squared straight y space dy space equals space integral cos space 2 straight x space dx

therefore space space space space space space space space space tan space straight y space equals space fraction numerator sin space 2 straight x over denominator 2 end fraction plus straight c

is the required solution.

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