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Differential Equations

Short Answer Type

211. Find a one parameter family of solutions of each of the following differential equation:
y² + x² y' = x y y'

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Long Answer Type

212. Find a one parameter family of solutions of each of the following differential equation:
(y² – 2xy) dx = (x² – 2xy) dy

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213. Find a one parameter family of solutions of each of the following differential equation:
y² dx + (x² – x y + y²) dy = 0

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214. Find a one parameter family of solutions of each of the following differential equation:
(x² + x y) dy = (x² + y²) dx

The given differential equation is
(x² + x y) dy = (x² + y²) dx
or

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

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

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

therefore space space space from space left parenthesis 1 right parenthesis comma space space space space straight v plus straight x dv over dx equals fraction numerator straight x squared plus straight v squared space straight x squared over denominator straight x squared plus straight v space straight x squared end fraction rightwards double arrow space space space space straight v plus straight x dv over dx equals fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction space space space space space space space rightwards double arrow space space space straight x dv over dx space equals fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction minus straight v therefore space space space space straight x dv over dx space equals space fraction numerator 1 minus straight v over denominator 1 plus straight v end fraction space space space space rightwards double arrow space space space space fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction dv space equals space 1 over straight x dx space space space space rightwards double arrow space space integral open parentheses negative 1 plus fraction numerator 2 over denominator 1 minus straight v end fraction close parentheses dv space equals integral 1 over straight x dx therefore space space minus straight v minus 2 space log space left parenthesis 1 minus straight v right parenthesis space equals space log space straight x space plus straight c space space space space rightwards double arrow space space space space space minus straight y over straight x minus 2 space log space open parentheses 1 minus straight y over straight x close parentheses space equals space log space straight x plus straight c therefore space space space space space straight y minus 2 straight x space log space open parentheses fraction numerator straight x minus straight y over denominator straight x end fraction close parentheses space equals space straight x space logx plus space cx

which is required solution.

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215. Solve the following differential equation:

open parentheses 1 plus straight e to the power of straight x over straight y end exponent close parentheses space dx space plus space straight e to the power of straight x over straight y end exponent space open parentheses 1 minus straight x over straight y close parentheses dy space equals space 0

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Short Answer Type

216.

Solve $open parentheses straight x space sin space straight y over straight x close parentheses dy space equals open parentheses straight y space sin space straight y over straight x minus straight x close parentheses dx$

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Long Answer Type

217.

Show that the differential equation $straight x space cos space open parentheses straight y over straight x close parentheses dy over dx space equals straight y space cos space open parentheses straight y over straight x close parentheses plus straight x$ is homogeneous and solve it.

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218.

Show that the differential equation:
$left parenthesis straight x space dy space minus space straight y space dx right parenthesis space straight y space sin space open parentheses straight y over straight x close parentheses space equals space left parenthesis straight y space dx space plus space straight x space dy right parenthesis space straight x space cos space open parentheses straight y over straight x close parentheses$

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Short Answer Type

219.

Solve:
$straight y space straight e to the power of straight x over straight y end exponent dx space equals open parentheses xe to the power of straight x over straight y end exponent plus straight y squared close parentheses space space dy comma space space space straight y not equal to 0$