For the following differential equation, find the general soluti

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Differential Equations

Short 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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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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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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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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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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126.

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

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127.

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

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128.
For the following differential equation, find the general solution:
$dy over dx equals fraction numerator 1 minus cosx over denominator 1 plus cosx end fraction$

The given differential equation is

dy over dx space equals space fraction numerator 1 minus cosx over denominator 1 plus cosx end fraction

or

dy space equals space fraction numerator 1 minus cosx over denominator 1 plus cosx end fraction dx space space space or space space space space integral dy space equals space fraction numerator 1 minus cosx over denominator 1 plus cosx end fraction dx

therefore space space space space space integral dy space equals space integral fraction numerator 2 space sin squared begin display style straight x over 2 end style over denominator 2 space cos squared space begin display style straight x over 2 end style end fraction dx space space or space space space integral space 1. space dy space equals space integral tan squared straight x over 2 dx

or

integral 1. space dy space equals space integral open parentheses sec squared straight x over 2 minus 1 close parentheses space dx comma space space or space space space straight y space equals space fraction numerator tan begin display style straight x over 2 end style over denominator begin display style 1 half end style end fraction minus straight x plus straight c

or

straight y space equals space 2 space tan straight x over 2 minus straight x plus straight c

which is the required solution.

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129.

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

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130. Solve 3e^x tan y dx + (1 + e^x) sec² y dy = 0.

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