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

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138.
Solve:
$left parenthesis 1 plus straight e to the power of 2 straight x end exponent right parenthesis space dy space plus space left parenthesis 1 plus straight y squared right parenthesis space straight e to the power of straight x space dx space equals space 0$

The given differential equation is

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 space space or space space left parenthesis 1 plus straight e to the power of 2 straight x end exponent right parenthesis space dy space equals space minus left parenthesis 1 plus straight y squared right parenthesis space straight e to the power of straight x space dx

Separating the variables, we get,

fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space minus fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx

Integrating,

integral fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space minus integral fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx

...(1)
Let

straight I space equals integral fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx

Put

straight e to the power of straight x space equals space straight t comma space space space space therefore space space space straight e to the power of straight x dx space equals space dt

therefore space space space space space space space space space straight I space equals space integral fraction numerator 1 over denominator 1 plus straight t squared end fraction dt space space equals space tan to the power of negative 1 end exponent straight t plus straight c subscript 1 space equals space tan to the power of negative 1 end exponent left parenthesis straight e to the power of straight x right parenthesis space plus space straight c subscript 1 therefore space space space space from space left parenthesis 1 right parenthesis comma space space space tan to the power of negative 1 end exponent straight y space equals space minus tan to the power of negative 1 end exponent left parenthesis straight e to the power of straight x right parenthesis space plus space straight c subscript 1 therefore space space space space tan to the power of negative 1 end exponent straight y space plus space tan to the power of negative 1 end exponent left parenthesis straight e to the power of straight x right parenthesis space equals space straight c subscript 1 therefore space space space tan to the power of negative 1 end exponent open parentheses fraction numerator straight y plus straight e to the power of straight x over denominator 1 minus ye to the power of straight x end fraction close parentheses space equals space straight c subscript 1 therefore space space space fraction numerator straight y plus straight e to the power of straight x over denominator 1 minus ye to the power of straight x end fraction space equals space straight c comma space space space where space straight c space equals space tan space straight c subscript 1

This is the required solution.

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139.

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

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140.

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