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Differential Equations

Long Answer Type

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

The given differential equation is
$dy over dx plus fraction numerator 4 straight x over denominator straight x squared plus 1 end fraction straight y space equals space minus fraction numerator 1 over denominator left parenthesis straight x squared plus 1 right parenthesis cubed end fraction$
Comparing it with $dy over dx plus straight P space straight y space equals space straight Q comma space space we space get comma space space space straight P space equals space fraction numerator 4 straight x over denominator straight x squared plus 1 end fraction comma space straight Q space equals negative fraction numerator 1 over denominator left parenthesis straight x squared plus 1 right parenthesis cubed end fraction$
$integral straight P space dx space equals space integral fraction numerator 4 straight x over denominator straight x squared plus 1 end fraction dx space equals space 2 integral fraction numerator 2 straight x over denominator straight x squared plus 1 end fraction dx space equals space 2 space log space left parenthesis straight x squared plus 1 right parenthesis space equals space log space left parenthesis straight x squared plus 1 right parenthesis squared integral straight e to the power of Pdx space equals space straight e to the power of log space left parenthesis straight x squared plus 1 right parenthesis squared end exponent equals space left parenthesis straight x squared plus 1 right parenthesis squared$
$therefore$ solution of given differential equation is
$straight y. space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dx end exponent space dx space plus straight c$
$or space space straight y. space left parenthesis straight x squared plus 1 right parenthesis squared space equals space integral fraction numerator negative 1 over denominator left parenthesis straight x squared plus 1 right parenthesis cubed end fraction. space left parenthesis straight x squared plus 1 right parenthesis squared space dx plus straight c or space space straight y. space open parentheses straight x squared plus 1 close parentheses squared space equals space minus integral fraction numerator 1 over denominator straight x squared plus 1 end fraction dx plus straight c space space or space space space straight y space left parenthesis straight x squared plus 1 right parenthesis squared space equals space minus tan to the power of negative 1 end exponent straight x plus straight c$

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Short 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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Long Answer Type

283.

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