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

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139.

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

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140.
Solve:
$straight x squared left parenthesis straight y minus 1 right parenthesis space dx space plus space straight y squared space left parenthesis straight x minus 1 right parenthesis space dy space equals space 0$

The given differential equation is
$straight x squared left parenthesis straight y minus 1 right parenthesis space dx space plus space straight y squared left parenthesis straight x minus 1 right parenthesis space dy space equals space 0 space space or space space straight y squared left parenthesis straight x minus 1 right parenthesis space dy space equals space minus straight x squared left parenthesis straight y minus 1 right parenthesis space dx$
or $fraction numerator straight y squared over denominator straight y minus 1 end fraction dy space equals space minus fraction numerator straight x squared over denominator straight x minus 1 end fraction dx$
Integrating, $integral fraction numerator straight y squared over denominator straight y minus 1 end fraction dy space equals space minus integral fraction numerator straight x squared over denominator straight x minus 1 end fraction dx$
$therefore space space space space space integral open parentheses straight y plus 1 plus fraction numerator 1 over denominator straight y minus 1 end fraction close parentheses dy space equals space minus integral open parentheses straight x plus 1 plus fraction numerator 1 over denominator straight x minus 1 end fraction close parentheses dx therefore space space straight y squared over 2 plus straight y plus log space open vertical bar straight y minus 1 close vertical bar space equals space minus open square brackets straight x squared over 2 plus straight x plus log space open vertical bar straight x plus 1 close vertical bar close square brackets plus straight c$
is the required solution.