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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.

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

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

The given differential equation is $dx over dy plus straight P subscript 1 straight x space equals space straight Q subscript 1$
Multiplying through by $straight e to the power of integral straight P subscript 1 dy end exponent comma space$ we get,
$dx over dy straight e to the power of integral straight P subscript 1 dy end exponent space plus space straight P subscript 1 straight x space straight e to the power of integral straight P subscript 1 dy end exponent space equals straight Q subscript 1 space straight e to the power of integral straight P subscript 1 end exponent space dy$
or $straight d over dy left parenthesis space straight x space straight e to the power of integral straight P subscript 1 dy end exponent right parenthesis space equals space straight Q subscript 1 space straight e to the power of integral straight P subscript 1 end exponent dy$
$open square brackets because space space straight d over dy left parenthesis straight x space straight e to the power of integral straight P subscript 1 dy end exponent right parenthesis space equals space xe to the power of integral straight P subscript 1 dy. end exponent straight d over dy left parenthesis integral straight P subscript 1 space dy right parenthesis plus straight e to the power of integral straight P subscript 1 dy end exponent dx over dy space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space xe to the power of integral straight P subscript 1 dy end exponent space space straight P subscript 1 space plus space straight e to the power of integral straight P subscript 1 dy end exponent space dx over dy space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space dx over dy straight e to the power of integral straight P subscript 1 dy end exponent space plus space straight P subscript 1 space xe to the power of integral straight P subscript 1 dy end exponent close square brackets space space space$
Integrating both sides w.r.t y, we get,
$xe to the power of integral straight P subscript 1 dy end exponent space equals space integral straight Q subscript 1 space straight e to the power of integral straight P subscript 1 dy end exponent dy plus straight C space which space is space required space solution. space$
$therefore space space space left parenthesis straight C right parenthesis space is space correct space answer. space$

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