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Differential Equations

Short Answer Type

191.

Solve the differential equation:
$straight x squared dy over dx space equals straight y left parenthesis straight x plus straight y right parenthesis.$

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Long Answer Type

192.

Solve the differential equation:
$left parenthesis straight y plus straight x right parenthesis space dy over dx space equals space straight y minus straight x.$

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193.

Solve the following differential equation:
(y² – x²) dy = 3 x y dx

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194. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(x - y) y' = x + 2 y

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195. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(x² + y²) y' = 8 x² - 3 x y + 2 y²

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196. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(3 x y + y²) dx = (x² + x y) dy

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

dy over dx equals fraction numerator 3 xy plus straight y squared over denominator straight x squared plus xy 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 3 straight x squared straight v plus straight v squared straight x squared over denominator straight x squared plus straight x squared straight v end fraction therefore space space space straight v plus straight x dv over dx space equals space fraction numerator 3 straight v plus straight v squared over denominator 1 plus straight v end fraction therefore space space space space space space straight x dv over dx space equals space fraction numerator 3 straight v plus straight v squared over denominator 1 plus straight v end fraction minus straight v space space space space space or space space space straight x dv over dx space equals space fraction numerator 3 straight v plus straight v squared minus straight v minus straight v squared over denominator 1 minus straight v end fraction therefore space space space space space space straight x dv over dx equals fraction numerator 2 straight v over denominator 1 plus straight v end fraction

Separating the variables,

fraction numerator 1 plus straight v over denominator straight v end fraction dv space equals space 2 over straight x dx

Integrating,

integral open parentheses 1 over straight v plus 1 close parentheses space dv space equals space 2 integral 1 over straight x dx

therefore space space space space space log space open vertical bar straight v close vertical bar plus straight v space equals space 2 space log space open vertical bar straight x close vertical bar space plus space straight c therefore space space space space log space open vertical bar straight y over straight x close vertical bar plus straight y over straight x space equals space 2 space log space open vertical bar straight x close vertical bar plus straight c

is the required solution.

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197. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
2 x y dx + (x² + 2 y²) dy = 0

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