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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:
$dy over dx equals straight e to the power of straight x minus straight y end exponent plus straight x squared straight e to the power of straight y$

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 x space space space space rightwards double arrow space space space space dy over dx space equals space straight e to the power of straight x. space straight e to the power of straight y plus space straight x squared. space straight e to the power of straight y.

rightwards double arrow space space space space space space space dy over dx space equals space left parenthesis straight e to the power of straight x plus straight x squared right parenthesis space straight e to the power of straight y

Separating the variables, we get,

dy over straight e to the power of negative straight y end exponent space equals space left parenthesis straight e to the power of straight x plus straight x squared right parenthesis dx space space rightwards double arrow space space space integral straight e to the power of straight y space dy space space equals space integral left parenthesis straight e to the power of straight x plus straight x squared right parenthesis space dx

therefore space space space space straight e to the power of straight y space equals space straight e to the power of straight x plus straight x cubed over 3 plus straight c

which is the required solution.

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