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Differential Equations

Short Answer Type

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

The given differential equation is 4 e^x tan y dx + 3(1 + e^x ) sec² y dy = 0.

therefore space space 3 space left parenthesis 1 plus straight e to the power of straight x right parenthesis space sec squared straight y space dy space equals negative space 4 space straight e to the power of straight x space tan space straight y space dx

fraction numerator sec squared straight y over denominator tan space straight y end fraction dy space equals space minus 4 over 3 fraction numerator straight e to the power of straight x over denominator 1 plus 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 space minus 4 over 3 integral fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of straight x end fraction dx

therefore space space log space open vertical bar tan space straight y close vertical bar space equals negative 4 over 3 space log space left parenthesis 1 plus straight e to the power of straight x right parenthesis space plus space log space straight c therefore space log space open vertical bar tan space straight y close vertical bar space equals space log space left parenthesis 1 plus straight e to the power of straight x right parenthesis to the power of fraction numerator negative 4 over denominator 3 end fraction end exponent space plus space log space straight c. space space space space space therefore space space space log space open vertical bar tan space straight y close vertical bar space equals space log space fraction numerator straight c over denominator left parenthesis 1 plus straight e to the power of straight x right parenthesis to the power of begin display style 4 over 3 end style end exponent end fraction therefore space space space space space open vertical bar tan space straight y close vertical bar space equals fraction numerator straight c over denominator left parenthesis 1 plus straight e to the power of straight x right parenthesis to the power of begin display style 4 over 3 end style end exponent end fraction space is space the space required space solution. space

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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.

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