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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:
$dy over dx plus straight y space cotx space equals space 4 straight x$
$cosec space straight x space left parenthesis straight x not equal to 0 right parenthesis comma space space given space that space straight y space equals space 0 space space space when space straight x space equals space straight pi over 2.$

The given differential equation is

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

dy over dx plus left parenthesis cotx right parenthesis. space straight y space equals space 4 space straight x space cosec space straight x

Comparing it with

dy over dx plus Py space equals space straight Q comma space we space get comma

straight P space space equals space cot space straight x comma space space space space straight Q space equals space 4 straight x space cosec space straight x

integral straight P space dx space equals space integral cotx space dx space equals space log space sinx space

therefore space space space space space space space space space space 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 log space sinx end exponent space equals space sin space straight x

therefore space

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

or space space space space space space space space straight y space sinx space equals space integral 4 straight x space cosecx. space sinx space dx space plus straight c or space space space space space space space space straight y space sinx space space equals space 4 space integral space straight x space dx space plus space straight c or space space space space space space space space straight y space sinx space space equals space 2 straight x squared plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis Now space straight y space equals space 0 space space space when space straight x space equals space straight pi over 2

therefore space space space space space space space space space space space space space space space space space 0 space equals space 2 cross times straight pi squared over 4 plus straight c space space space space space space space space space space space space space space rightwards double arrow space space space space space space space space straight c space equals space minus straight pi squared over 2

therefore

from (1),

straight y space sinx space equals space 2 straight x squared minus straight pi squared over 2

which is required solution.

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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.

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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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