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Differential Equations

Short Answer Type

131. Solve 4 e^x tan y dx + 3(1 + e^x) sec²y dy = 0.

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Long Answer Type

132.

For the following differential equation, find the general solution:
sec² x tan y dx – sec² y tan x dy = 0.

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Short Answer Type

133.

For the following differential equation, find the general solution:
sec² x tan y dx + sec² y tan x dy = 0.

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134.
Solve:
3e^x tan y dx + (1 – e^x) sec²y dy = 0.

The given differential equation is
3e^x tan y dx + (1 – e^x) sec²y dy = 0
or $left parenthesis 1 minus straight e to the power of straight x right parenthesis space sec squared straight y space dy space equals negative 3 space straight e to the power of straight x space tan space straight y space dx space space space or space space space fraction numerator sec squared straight y over denominator tan space straight y end fraction dy space equals space minus 3 fraction numerator straight e to the power of straight x over denominator 1 minus straight e to the power of straight x end fraction dx$
$therefore space space space space integral fraction numerator sec squared straight y over denominator tan space straight y end fraction dy space equals 3 integral fraction numerator straight e to the power of negative straight x end exponent over denominator 1 minus straight e to the power of straight x end fraction dx space space space space space space space rightwards double arrow space space space space log space left parenthesis tan space straight y right parenthesis space equals space 3 space log space left parenthesis 1 minus straight e to the power of straight x right parenthesis space plus space log space straight c$
$therefore space space space log space left parenthesis tan space straight y right parenthesis space equals space log space left parenthesis 1 minus straight e to the power of straight x right parenthesis cubed plus space log space straight c space space space space rightwards double arrow space space space log space left parenthesis tany right parenthesis space equals space log space open square brackets straight e left parenthesis 1 minus straight e to the power of straight x right parenthesis cubed close square brackets therefore space space space tan space straight y space equals space straight e left parenthesis 1 minus straight e to the power of straight x right parenthesis cubed space is space the space required space solution. space$

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135.

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

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136.

Solve:
tan y dx + sec² y tan x dy = 0.

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137.

Solve:
$left parenthesis 1 plus straight y squared right parenthesis thin space left parenthesis 1 plus log space straight x right parenthesis space dx space plus space straight x space dy space equals space 0$

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138.

Solve:
$left parenthesis 1 plus straight e to the power of 2 straight x end exponent right parenthesis space dy space plus space left parenthesis 1 plus straight y squared right parenthesis space straight e to the power of straight x space dx space equals space 0$

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139.

Solve:
$log dy over dx equals ax plus by.$

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140.

Solve:
$straight x squared left parenthesis straight y minus 1 right parenthesis space dx space plus space straight y squared space left parenthesis straight x minus 1 right parenthesis space dy space equals space 0$