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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.

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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.

Let y = f (x) be the equation of curve.
We know that

dy over dx

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

dy over dx space equals space straight x plus straight y

dy over dx minus straight y space equals straight x

Comparing

dy over dx plus straight P space straight y space equals space straight Q space space space with space dy over dx minus straight y space equals space straight x comma space we space get comma space space straight P space equals space minus 1 comma space space straight Q space equals space straight x

therefore space space space space space space space space integral straight P space dx space equals space minus integral 1 space dx space equals space minus straight x 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 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 integral straight Q space straight e to the power of integral straight P space dx end exponent dx space plus space straight c

straight y space straight e to the power of negative straight x end exponent space equals space integral space straight x space straight e to the power of negative straight x end exponent space dx plus straight c space space space or space space space space straight y space straight e to the power of negative straight x end exponent space equals space straight x fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction minus integral 1. space fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction dx plus straight c

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

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

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

straight x plus straight y plus 1 space equals space straight c space straight e to the power of straight x

...(1)
Now curve passes through origin (0, 0).

therefore space space space 0 plus 0 plus 1 space equals space straight c space straight e to the power of 0 space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight c space equals space 1

therefore

from (1),

straight x plus straight y plus 1 space equals space straight e to the power of straight x

is the required equation of curve.

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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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