The differential equation of y = aebx (a and b are parameter

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 Multiple Choice QuestionsShort Answer Type

31.

Find the general solution of (x + log(y))dy + ydx = 0


 Multiple Choice QuestionsMultiple Choice Questions

32.

The general solution of the differential equation

d2ydx2 +8dydx + 16y = 0 is

  • (A + B)e5x

  • (A + Bx)e- 4x

  • (A + Bx2)e4x

  • (A + Bx4)e4x


33.

If x2 + y2 = 4, then ydydx + x is equal to

  • 4

  • 0

  • 1

  • - 1


34.

x3dx1 + x8 is equal to

  • 4tan-1x4 + C

  • 14tan-1x3 + C

  • x +4tan-1x4 + C

  • x2 +14tan-1x4 + C


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35.

The degree and order of the differential equation

y = xdydx2 + dxdy2 are respectively

  • 1, 1

  • 2, 1

  • 4, 1

  • 1, 4


36.

The general solution of the differential equation logedydx = x + y is

  • ex + e- y = C

  • ex + ey = C

  • ey + e- x = C

  • e- x + e- y = C


37.

If y = Ax + Bx2, then x2d2ydx2 is equal to

  • 2y

  • y2

  • y3

  • y4


38.

The solution of dydx = yx + tanyx is

  • x = csinyx

  • x = csinxy

  • y = csinyx

  • xy = csinxy


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39.

Integrating Factor (IF) of the differential equation dydx - 3x2y1 + x3 = sin2x1 + x

  • e1 + x3

  • log1 + x3

  • 1 + x3

  • 11 + x3


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40.

The differential equation of y = aebx (a and b are parameters) is

  • yy1 = y22

  • yy2 = y12

  • yy12 = y2

  • yy22 = y1


B.

yy2 = y12

Given, y = aebx

On differentiating w.r.t. x, we get

     y1 = abebx

 y1 = by         ...(i)

Again differentiating, we get

    y2 = by1

  y2 = y1y . y1      from Eq. (i) y12 = yy2


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