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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.

90 Views

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.

75 Views

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

The given differential equation is
tan y dx + sec² y tan x dy = 0 or sec² y tan x dy = - tan y dx
Separating the variables, we get,

fraction numerator sec squared straight y over denominator tan space straight y end fraction dy space equals space minus fraction numerator 1 over denominator tan space straight x end fraction dx

Integrating,

integral fraction numerator sec squared straight y over denominator tan space straight y end fraction dy space equals negative fraction numerator 1 over denominator tan space straight x end fraction dx

therefore space space log space open vertical bar tan space straight y close vertical bar space equals space minus log space open vertical bar sin space straight x close vertical bar space plus space log space straight c apostrophe therefore space space log space open vertical bar tan space straight y close vertical bar space plus space log space open vertical bar sin space straight x close vertical bar space equals space log space straight c apostrophe therefore space space log space open vertical bar tan space straight y space sinx space close vertical bar space equals space log space straight c apostrophe therefore space space open vertical bar tany space sinx close vertical bar space equals space straight c apostrophe therefore space space tany space sinx space space equals space straight A comma space where space straight A space is space constant. space

126 Views

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

The given differential equation is
$straight e to the power of straight x space tany space dx space plus space left parenthesis 1 minus straight e to the power of straight x right parenthesis space sec squared straight y space dy space equals space 0$
or $left parenthesis 1 minus straight e to the power of straight x right parenthesis space sec squared straight y space dy space equals space minus straight e to the power of straight x space tan space straight y space dx space space space or space space space fraction numerator sec squared straight y over denominator tan space straight y end fraction space dy space equals space minus fraction numerator straight e to the power of straight x over denominator 1 minus straight e to the power of straight x end fraction dx$
$therefore space space space integral fraction numerator sec squared straight y over denominator tan space straight y end fraction dy space equals space integral fraction numerator negative straight e to the power of straight x over denominator 1 minus straight e to the power of straight x end fraction dx rightwards double arrow space space space space log space left parenthesis tan space straight y right parenthesis space equals space log space left parenthesis 1 minus straight e to the power of straight x right parenthesis space plus space log space straight e rightwards double arrow space space space space log space left parenthesis tan space straight y right parenthesis space equals space log space left square bracket straight e space left parenthesis 1 minus straight e to the power of straight x right parenthesis right square bracket therefore space space space space space space space tan space straight y space equals space straight e left parenthesis 1 minus straight e to the power of straight x right parenthesis space is space the space required space solution. space space$

83 Views

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