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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

The given differential equation is
y² dx + (x² – x y + y²) dy = 0 or (x² – x y + y²) dy = - y² dx

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

Put y = vx 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 negative fraction numerator straight v squared straight x squared over denominator straight x squared minus vx squared plus straight v squared straight x squared end fraction therefore space space space space space straight v plus straight x dv over dx equals negative fraction numerator straight v squared over denominator 1 minus straight v plus straight v squared end fraction therefore space space space space space space straight x dv over dx space equals space minus fraction numerator straight v squared over denominator 1 minus straight v plus straight v squared end fraction minus straight v therefore space space space space space straight x dv over dx equals fraction numerator negative straight v squared minus straight v plus straight v squared minus straight v cubed over denominator 1 minus straight v plus straight v squared end fraction therefore space space space space space space straight x dv over dx equals fraction numerator negative straight v minus straight v cubed over denominator 1 minus straight v plus straight v squared end fraction therefore space space fraction numerator 1 minus straight v plus straight v squared over denominator straight v plus straight v cubed end fraction dv space equals space minus 1 over straight x dx therefore space space space space space integral fraction numerator 1 minus straight v plus straight v squared over denominator straight v left parenthesis 1 plus straight v squared right parenthesis end fraction space identical to space straight A over straight v plus fraction numerator Bv plus straight c over denominator 1 plus straight v squared end fraction

Multiplying both sides by v (1 + v²), we get.

1 minus straight v plus straight v squared space equals space straight A thin space left parenthesis 1 plus straight v squared right parenthesis space plus space Bv squared plus Cv

...(1)
Put v = 0 in (1)

therefore space space space space space space 1 minus 0 plus 0 space equals space straight A space left parenthesis 1 plus 0 right parenthesis space plus space 0 space plus space 0. space space space space space space rightwards double arrow space space space straight A space equals space 1

Equating coefficients in (1) of

straight v squared right parenthesis

1 = A + B

rightwards double arrow 1 space equals 1 space plus space straight B space space space space space space space space space space space space space space space space rightwards double arrow space space space straight B space equals space 0

v) -1 = C

rightwards double arrow space straight C space equals space minus 1

therefore space space space space space fraction numerator 1 minus straight v plus straight v squared over denominator straight v left parenthesis 1 plus straight v squared right parenthesis end fraction identical to 1 over straight v plus fraction numerator negative 1 over denominator 1 plus straight v squared end fraction

therefore space space space space space from space left parenthesis 1 right parenthesis comma space we space get

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

therefore space space log space open vertical bar straight v close vertical bar space minus space tan to the power of negative 1 end exponent straight v space equals space space minus space log space open vertical bar straight x close vertical bar space plus space straight c therefore space space space log space open vertical bar straight y over straight x close vertical bar space minus space tan to the power of negative 1 end exponent straight y over straight x space equals space minus log space open vertical bar straight x close vertical bar space plus space log space straight A therefore space space space log space open vertical bar straight y close vertical bar space minus space log space open vertical bar straight x close vertical bar space minus space tan to the power of negative 1 end exponent straight y over straight x space equals space minus log space open vertical bar straight x close vertical bar plus space space log space straight A therefore space space space log space open vertical bar straight y close vertical bar minus log space straight A space equals space space tan to the power of negative 1 end exponent straight y over straight x therefore space space space log space open vertical bar fraction numerator open vertical bar straight y close vertical bar over denominator straight A end fraction close vertical bar space equals tan to the power of negative 1 end exponent straight y over straight x space space space space rightwards double arrow space space space space fraction numerator open vertical bar straight y close vertical bar over denominator straight A end fraction space equals space straight e to the power of tan to the power of negative 1 end exponent straight y over straight x end exponent therefore space space space space space space space space space space space space space space space space space space open vertical bar straight y close vertical bar space equals space Ae to the power of tan to the power of negative 1 end exponent straight y over straight x end exponent therefore space space space space space space space space space space space space space space space space space space straight y squared space equals space space straight c space straight e space to the power of 2 space tan to the power of negative 1 end exponent straight y over straight x end exponent

is the required solution.

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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

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