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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.

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

The given differential equation is
$left parenthesis 1 plus straight y squared right parenthesis thin space left parenthesis 1 plus log space straight x right parenthesis space dx space plus space straight x space dy space equals space 0 space space or space space straight x space dy space equals space minus left parenthesis 1 plus straight y squared right parenthesis thin space left parenthesis 1 plus space log space straight x right parenthesis space dx$
Separating the variables, we get,
$fraction numerator dy over denominator 1 plus straight y squared end fraction space equals negative fraction numerator 1 plus log space straight x over denominator straight x end fraction dx$
Integrating, $integral fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space minus integral left parenthesis 1 plus log space straight x right parenthesis. space 1 over straight x dx$
$therefore space space space tan to the power of negative 1 end exponent straight y space plus straight c space equals space minus fraction numerator left parenthesis 1 plus space log space straight x right parenthesis squared over denominator 2 end fraction space space space space space open square brackets because space space integral open square brackets straight f left parenthesis straight x right parenthesis close square brackets to the power of straight n space straight f apostrophe left parenthesis straight x right parenthesis space dx space equals space fraction numerator straight f to the power of straight n plus 1 end exponent left parenthesis straight x right parenthesis over denominator straight n plus 1 end fraction plus straight c close square brackets therefore space space space tan to the power of negative 1 end exponent straight y space plus space 1 half left parenthesis 1 plus space log space straight x right parenthesis squared space plus space straight c space equals space 0$
which is the required solution.

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