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Differential Equations

Long Answer Type

291. For the given differential equation, find a particular solution satisfying the given condition:

left parenthesis 1 plus straight x squared right parenthesis space dy over dx space plus space 2 space straight x space straight y space space equals space fraction numerator 1 over denominator 1 plus straight x squared end fraction semicolon space straight y space equals space 0 space space when space straight x space equals space 1

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292. For the given differential equation, find a particular solution satisfying the given condition:

dy over dx minus 3 space straight y space cotx space equals space sin space 2 straight x space colon space space straight y space equals space 2 space space when space space straight x space equals space straight pi over 2

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293. Find a particular solution of the differential equation:

dy over dx plus straight y space cotx space equals space 4 straight x

cosec space straight x space left parenthesis straight x not equal to 0 right parenthesis comma space space given space that space straight y space equals space 0 space space space when space straight x space equals space straight pi over 2.

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294. Find a particular solution of the differential equation

left parenthesis straight x plus 1 right parenthesis space dy over dx space equals space 2 straight e to the power of negative straight y end exponent minus 1 comma space space

given that y = 0, when x = 0.

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295. Solve the differential equation, given that y = 1 where x = 2

straight x dy over dx plus straight y space equals space straight x cubed.

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296. Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of x coordinate (abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point.

Let y = f (x) be equation of curve.
Now

dy over dx

is the slope of the tangent to the curve at point (x, y)
From the given condition,

dx over dy space equals space straight x plus xy

dy over dx minus straight x space straight y space equals space straight x

Comparing

dy over dx plus Py space equals space straight Q

with

dy over dx minus straight x space straight y space equals space straight x comma space

we get, P = -x, Q = x

therefore space space space space space space space integral straight P space dx space equals negative integral straight x space dx space equals space minus straight x squared over 2 comma space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of negative straight x squared over 2 end exponent

Solution of differential equation is

straight y space straight e to the power of integral straight P space dx end exponent space equals space integral straight Q space straight e to the power of integral straight P space dx end exponent dx plus straight c

straight y space straight e to the power of negative straight x squared over 2 end exponent space equals space integral space straight x space straight e to the power of negative straight x squared over 2 end exponent dx plus straight c

...(1)
Let

straight I space equals space integral space straight x space space straight e to the power of negative straight x squared over 2 end exponent dx plus straight c

Put

negative straight x squared over 2 space equals space straight t comma space space space space therefore space space space space space space minus straight x space dx space space equals space dt space space space space rightwards double arrow space space space space straight x space dx space equals space minus dt

therefore

straight I space equals space minus integral straight e to the power of straight t dt space equals space minus straight e to the power of straight t space equals space minus straight e to the power of negative straight x squared over 2 end exponent

therefore

straight y space straight e to the power of negative straight x squared over 2 end exponent space equals space minus straight e to the power of negative straight x squared over 2 end exponent plus straight c

straight y space equals negative 1 plus space straight c space straight e to the power of straight x squared over 2 end exponent

...(2)
Since the curve passes through (0, 1)

therefore

straight I space equals space minus 1 plus space straight c space straight e to the power of 0 space space space space rightwards double arrow space space space 2 space equals space straight c

therefore

from (2),

straight y equals negative 1 plus 2 space straight e to the power of straight x squared over 2 end exponent

straight y plus 1 space equals space 2 space straight e to the power of straight x squared over 2 end exponent

, which is required equation of curve.

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297. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

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Short Answer Type

298.

The integrating factor of the differential equation $straight x dy over dx minus straight y space equals space 2 straight x squared$ is

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Multiple Choice Questions

299.

The integrating factor of the differential equation:
$left parenthesis 1 minus straight y squared right parenthesis dx over dy plus straight y space straight x space equals space straight a space straight y space space left parenthesis negative 1 less than straight y less than 1 right parenthesis$

$fraction numerator 1 over denominator straight y squared minus 1 end fraction$
$fraction numerator 1 over denominator square root of straight y squared minus 1 end root end fraction$
$fraction numerator 1 over denominator 1 minus straight y squared end fraction$
$fraction numerator 1 over denominator 1 minus straight y squared end fraction$

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300. The general solution of a differential equation of the type

dx over dy plus straight P subscript 1 straight x space equals space straight Q subscript 1

$straight y space straight e to the power of integral straight P subscript 1 dy end exponent space equals space integral space open parentheses straight Q subscript 1 space straight e to the power of integral straight P subscript 1 dy end exponent close parentheses space dy plus straight C$
$straight y. space straight e to the power of integral straight P subscript 1 dx end exponent space equals space integral space open parentheses straight Q subscript 1 space straight e to the power of integral straight P subscript 1 dx end exponent close parentheses dx plus straight C$
$straight x. straight e to the power of integral straight P subscript 1 dy end exponent space equals space integral open parentheses straight Q subscript 1 straight e to the power of integral straight P subscript 1 dy end exponent close parentheses dy plus straight C$
$straight x. straight e to the power of integral straight P subscript 1 dy end exponent space equals space integral open parentheses straight Q subscript 1 straight e to the power of integral straight P subscript 1 dy end exponent close parentheses dy plus straight C$

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