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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

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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.

The given differential equation is

space straight x space cos space open parentheses straight y over straight x close parentheses space dy over dx space equals space straight y space cos space open parentheses straight y over straight x close parentheses plus space straight x space space space or space space space space dy over dx space equals fraction numerator straight y space cos space open parentheses begin display style straight y over straight x end style close parentheses plus straight x over denominator straight x space cos space open parentheses begin display style straight y over straight x end style close parentheses end fraction

...(1)
It is differential equation of the form

dy over dx space equals space straight F left parenthesis straight x comma space straight y right parenthesis.

Here,

straight F left parenthesis straight x comma space straight y right parenthesis space equals space fraction numerator straight y space cos space open parentheses begin display style straight y over straight x end style close parentheses plus straight x over denominator straight x space cos space open parentheses begin display style straight y over straight x end style close parentheses end fraction

Replacing x by

straight lambda space straight x

and y by

λy comma

we get

space straight F left parenthesis λx comma space λy right parenthesis space equals space fraction numerator straight lambda space open square brackets straight y space cos space open parentheses begin display style straight y over straight x end style close parentheses plus straight x close square brackets over denominator straight lambda space open parentheses straight x space cos space begin display style straight y over straight x end style close parentheses end fraction space equals space straight lambda degree space space left square bracket straight F space left parenthesis straight x comma space straight y right parenthesis right square bracket

therefore space space space straight F left parenthesis straight x comma space straight y right parenthesis

is a homogeneous function of degree zero.

therefore space

the given differential equation is a homogeneous differential equation
Put y = vx 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 straight v plus straight x dv over dx space equals space fraction numerator straight v space straight x space cos space open parentheses begin display style vx over straight x end style close parentheses plus straight x over denominator straight x space cos space open parentheses begin display style vx over straight x end style close parentheses end fraction

therefore space space space straight v plus straight x dv over dx space equals space fraction numerator straight v space straight x space cos space straight v space plus space straight x over denominator straight x space cos space straight v end fraction

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

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

Separating the variables, we get,

cosv space dv space equals space 1 over straight x dx

Integrating,

integral space cos space straight v space dv space equals space integral 1 over straight x dx

therefore

sin space straight v space equals space log space open vertical bar straight x close vertical bar space plus space log space open vertical bar straight c close vertical bar

sin space straight y over straight x space equals space log space open vertical bar straight c space straight x close vertical bar space which space is space required space solution. space

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