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Differential Equations

Long Answer Type

231.

Solve the following differential equation:
$dy over dx equals fraction numerator straight x left parenthesis 2 straight y minus straight x right parenthesis over denominator straight x left parenthesis 2 straight y plus straight x right parenthesis end fraction comma space space if space space straight y space equals space 1 space space when space straight x space equals space 1.$

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232.

Solve the following initial value problem:
$dy over dx space equals space fraction numerator straight y left parenthesis straight x plus 2 straight y right parenthesis over denominator straight x left parenthesis 2 straight x plus straight y right parenthesis end fraction. space space space straight y left parenthesis 1 right parenthesis space equals space 2$

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Short Answer Type

233. Show that the family of curves for which the slope of the tangent at any point (x, y) on it is $fraction numerator straight x squared plus straight y squared over denominator 2 xy end fraction comma$ is given by $straight x squared minus straight y squared space equals space straight c space straight x.$

We know that $dy over dx$ is slope of tangent to the curve at point (x, y).
$therefore space space space space space space space space space dy over dx space equals space fraction numerator straight x squared plus straight y squared over denominator 2 xy end fraction$
Put y = v x so that $dy over dx equals straight v plus straight x dv over dx$
$therefore space space space space straight v plus straight x dv over dx equals fraction numerator straight x squared plus straight v squared straight x squared over denominator 2 vx squared end fraction therefore space space space space straight v plus straight x dv over dx space equals fraction numerator 1 plus straight v squared over denominator 2 straight v end fraction therefore space space space space space space space space space straight x dv over dx space equals space fraction numerator 1 plus straight v squared over denominator 2 straight v end fraction minus straight v space space space space space space or space space space space space straight x dv over dx equals space fraction numerator 1 minus straight v squared over denominator 2 straight v end fraction$
Separating the variables and integrating, we get,
$integral fraction numerator 2 straight v over denominator 1 minus straight v squared end fraction dv space equals space integral 1 over straight x dx space space space space or space space space integral fraction numerator 2 straight v over denominator straight v squared minus 1 end fraction dv space equals space minus integral 1 over straight x dx$

$therefore space space space log space open vertical bar straight v squared minus 1 close vertical bar space equals space minus log space open vertical bar straight x close vertical bar plus log space open vertical bar straight c subscript 1 close vertical bar space space space or space space space log space open vertical bar left parenthesis straight v squared minus 1 right parenthesis thin space left parenthesis straight x right parenthesis close vertical bar space equals space log space open vertical bar straight c subscript 1 close vertical bar or space space space space space space space left parenthesis straight v squared minus 1 right parenthesis space straight x space equals space plus-or-minus space space straight c subscript 1$
Replacing v by $straight y over straight x comma space$ we get

$open parentheses straight y squared over straight x squared minus 1 close parentheses space straight x space equals space plus-or-minus space straight c subscript 1 space space space or space space space space left parenthesis straight y squared minus straight x squared right parenthesis space equals space space plus-or-minus space straight c subscript 1 straight x space space space or space space space straight x squared minus straight y squared space equals space cx.$

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Multiple Choice Questions

234. The general solution of the differential equation

fraction numerator straight y space dx space minus space straight x space dy over denominator straight y end fraction space equals space 0

xy = C
x = Cy²
y = Cx
y = Cx

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235.

A homogeneous differential equation of the from $dx over dy space equals space straight h open parentheses straight x over straight y close parentheses$ can be solved by making the substitution.

y = vx
v = yx
x = vy
x = vy

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236. Which of the following is a homogeneous differential equation?

(4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
(xy) dx – (x³ + y³) dy = 0
(x³ + 2 y²) dx + 2xy dy = 0
(x³ + 2 y²) dx + 2xy dy = 0

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Short Answer Type

237.

Solve : $dy over dx plus straight y space equals space sin space straight x comma space space left parenthesis straight x space element of space straight R right parenthesis$

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Long Answer Type

238. Solve the differential equation:

straight x dy over dx minus straight y minus 2 straight x cubed space equals space 0.

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239. Solve the differential equation:

dy over dx minus 2 straight y space equals space 3 straight x.

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Short Answer Type

240.

Solve $dy over dx plus straight y over straight x space equals space straight e to the power of straight x comma space space left parenthesis straight x greater than 0 right parenthesis.$