For the given differential equation, find the particular solut

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

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224.

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

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225.

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

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226.

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

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227.

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

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228.

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

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229.
For the given differential equation, find the particular solution satisfying the given condition:
$2 xy plus straight y squared minus 2 straight x squared straight y apostrophe space equals space 0 space space space space space space space space space space space space space straight y left parenthesis 1 right parenthesis space equals space 2$

The given differential equation is

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 or space space space space space 2 straight x squared straight y apostrophe space equals space 2 xy plus straight y squared

therefore space space space space space space space space space dy over dx space equals space fraction numerator 2 xy plus straight y squared over denominator 2 straight x squared end fraction

Put y - v x so that

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

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

Integration,

2 space integral straight v to the power of negative 2 end exponent space dv space equals space integral 1 over straight x dx

therefore space space space space space 2 fraction numerator straight v to the power of negative 1 end exponent over denominator negative 1 end fraction space equals space log space open vertical bar straight x close vertical bar plus straight c space space space space space or space space space space fraction numerator negative 2 over denominator straight v end fraction equals log space open vertical bar straight x close vertical bar plus straight c therefore space space space space minus 2 straight x over straight y space equals space log space open vertical bar straight x close vertical bar plus straight c

Now,

straight y left parenthesis 1 right parenthesis space equals space 2 space space space space space space space space space space space space space space rightwards double arrow space space space space straight y space equals space 2 space space space when space straight x space equals space 1

therefore space space space space minus 2. space 1 half space equals space log space 1 plus straight c space space space space space space space space space space rightwards double arrow space space space space minus 1 space equals space 0 space plus straight c space space space space rightwards double arrow space space space space straight c space equals space minus 1 therefore space space solution space is space minus fraction numerator 2 straight x over denominator straight y end fraction space equals space log space open vertical bar straight x close vertical bar minus 1 space space space space space or space space space straight y space equals space fraction numerator 2 straight x over denominator 1 minus log space open vertical bar straight x close vertical bar end fraction.

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Short Answer Type

230.

Find a particular solution of the differential equation
(x – y) (dx + dy) = dx – dy. given that y = – 1, when x = 0.

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