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Differential Equations

Short Answer Type

221.

Solve:
$straight y apostrophe space equals space straight y over straight x plus sin straight y over straight x$

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222.

Solve:
$dy over dx space equals straight y over straight x plus tan straight y over straight x open parentheses 0 less than straight y over straight x less than straight pi over 2 close parentheses$

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Long Answer Type

223.
For the given differential equations, find the particular solution satisfying the given condition:
$left parenthesis straight x plus straight y right parenthesis space dy space plus space left parenthesis straight x minus straight y right parenthesis dx space equals space 0 space semicolon space space straight y space equals space 1 space space space when space straight x space equals space 1$

The given differential equation is

left parenthesis straight x plus straight y right parenthesis space dy space plus space left parenthesis straight x minus straight y right parenthesis space dx space equals space 0 comma space space space or space space space left parenthesis straight x plus straight y right parenthesis space dy space equals space minus left parenthesis straight x minus straight y right parenthesis space dx

therefore space space space space space dy over dx space equals space minus fraction numerator straight x minus straight y over denominator straight x plus straight y end fraction space space space or space space space dy over dx space equals space fraction numerator straight y minus straight x over denominator straight y plus straight x end fraction

...(1)
Put y = v x, so that

dy over dx space equals straight v plus straight x dv over dx

therefore space space space space space from space left parenthesis 1 right parenthesis comma space space space space straight v plus straight x dv over dx space equals space fraction numerator straight v space straight x space minus straight x over denominator vx plus straight x end fraction space space or space space straight v plus straight x dv over dx space equals space fraction numerator straight v minus 1 over denominator straight v plus 1 end fraction

straight x dy over dx space equals fraction numerator straight v minus 1 over denominator straight v plus 1 end fraction minus straight v space equals space fraction numerator negative 1 minus straight v squared over denominator straight v plus 1 end fraction space equals space fraction numerator negative left parenthesis 1 plus straight v squared right parenthesis over denominator 1 plus straight v end fraction space space or space space fraction numerator 1 plus straight v over denominator 1 plus straight v squared end fraction dv space equals negative dx over straight x

integral fraction numerator 1 plus straight v over denominator 1 plus straight v squared end fraction dv space equals space minus integral dx over straight x

integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv plus 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space minus integral 1 over straight x dx

tan to the power of negative 1 end exponent straight v space plus space 1 half log space open vertical bar 1 plus straight v squared close vertical bar plus log space open vertical bar straight x close vertical bar space equals space straight c

tan to the power of negative 1 end exponent straight v plus 1 half left square bracket log space open vertical bar 1 plus straight v squared close vertical bar space plus space 2 space log space open vertical bar straight x close vertical bar right square bracket space equals space straight c

tan to the power of negative 1 end exponent straight v plus 1 half log space open vertical bar straight x squared close vertical bar space space open vertical bar 1 plus straight v squared close vertical bar space equals space straight c

or space space tan space to the power of negative 1 end exponent straight y over straight x space plus space 1 half space log space open vertical bar straight x squared plus open parentheses 1 plus straight y squared over straight x squared close parentheses close vertical bar space equals space straight c or space space space tan space to the power of negative 1 end exponent straight y over straight x plus 1 half log space open vertical bar straight x squared plus straight y squared close vertical bar space equals space straight c therefore space space space space tan to the power of negative 1 end exponent straight y over straight x plus 1 half log space left parenthesis straight x squared plus straight y squared right parenthesis space equals space straight c space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

Now y = 1 when x = 1

therefore space space space space space space tan to the power of negative 1 end exponent 1 plus 1 half log 2 space equals space straight c space space space space space rightwards double arrow space space space space space straight c space equals space straight pi over 4 plus 1 half log space 2 therefore space space space space from space left parenthesis 2 right parenthesis comma space space tan to the power of negative 1 end exponent straight y over straight x plus 1 half log space left parenthesis straight x squared plus straight y squared right parenthesis space equals space straight pi over 4 plus 1 half log space 2

is the required solution.

75 Views

224.

For the given differential equation, find the particular solution satisfying the given condition:
$straight x squared dy plus left parenthesis xy plus straight y squared right parenthesis dx space equals space 0 semicolon space space space straight y space equals space 1 space when space straight x space equals space 1$

80 Views

225.

For the given differential equation, find the particular solution satisfying the given condition:
$open square brackets straight x space sin squared open parentheses straight y over straight x close parentheses minus straight y close square brackets space dx plus space straight x space dy space equals space 0 colon space space space straight y space equals straight pi over 4 space space when space straight x space equals space 1$

73 Views

226.

For the given differential equation, find the particular solution satisfying the given condition:
$dy over dx minus straight y over straight x plus cosec space open parentheses straight y over straight x close parentheses space equals space 0 space semicolon space space space straight y space equals space 0 space space space when space straight x space equals space 1$

75 Views

227.

For the given differential equation, find the particular solution satisfying the given condition:
$2 xy plus straight y squared minus 2 straight x squared dy over dx equals 0 semicolon space space space straight y space equals 2 space space when space straight x space equals space 1.$

75 Views

228.

For the given differential equation, find the particular solution satisfying the given condition:
$2 straight x squared straight y apostrophe space minus space 2 xy plus straight y squared space equals space 0 comma space space space space straight y left parenthesis straight e right parenthesis space equals space straight e$

74 Views

229.

For the given differential equation, find the particular solution satisfying the given condition:
$2 xy plus straight y squared minus 2 straight x squared straight y apostrophe space equals space 0 space space space space space space space space space space space space space straight y left parenthesis 1 right parenthesis space equals space 2$