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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

The given differential equation is
(x y² + 2x) dx + (x² y + 2y) dy = 0
or $straight y left parenthesis straight x squared plus 2 right parenthesis space dx space equals space minus straight x left parenthesis straight y squared plus 2 right parenthesis space dx$
$therefore space space space space space fraction numerator straight y over denominator straight y squared plus 2 end fraction dy space equals negative fraction numerator straight x over denominator straight x squared plus 2 end fraction dx$
Integrating, $integral fraction numerator 2 straight y over denominator straight y squared plus 2 end fraction dy space equals negative integral fraction numerator 2 straight x over denominator straight x squared plus 2 end fraction dx$
$therefore space space space space space space log space left parenthesis straight y squared plus 2 right parenthesis space equals space minus log left parenthesis straight x squared plus 2 right parenthesis space plus space log space straight c space space space or space space space log space open square brackets left parenthesis straight y squared plus 2 right parenthesis space left parenthesis straight x squared plus 2 right parenthesis close square brackets space plus space log space straight c therefore space space space space left parenthesis straight y squared plus 2 right parenthesis thin space left parenthesis straight x squared plus 2 right parenthesis space equals space straight c space is space the space required space solution.$

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