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Differential Equations

Short Answer Type

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

The given differential equation is
$dy over dx space equals space fraction numerator straight y left parenthesis straight x squared minus 1 right parenthesis over denominator straight x left parenthesis straight y squared minus 1 right parenthesis end fraction$
Separating the variables, we get,
$fraction numerator straight y squared minus 1 over denominator straight y end fraction dy space equals space fraction numerator straight x squared minus 1 over denominator straight x end fraction dx$
$therefore space space space space space space integral open parentheses straight y squared over straight y minus 1 over straight y close parentheses dy space equals space integral open parentheses straight x squared over straight x minus 1 over straight x close parentheses dx$
or $integral open parentheses straight y minus 1 over straight y close parentheses dy space equals space integral open parentheses straight x minus 1 over straight x close parentheses dx$
or $straight y squared over 2 minus log space open vertical bar straight y close vertical bar space equals space straight x squared over 2 minus log space open vertical bar straight x close vertical bar space plus space straight c$
is the required solution.