The solution of tanydydx = sinx +&n

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 Multiple Choice QuestionsMultiple Choice Questions

361.

The solution of the differential equation dydx = sinx +ytanx + y - 1 is

  • cscx +y + tanx +y = x + c

  • x +cscx + y = c

  • x +tanx +y = c

  • x +secx +y = c


362.

The differential equation of the family  y = aex + bxex + cx2ex of curves, where a, b, c are arbitrary constants , is 

  • y''' + 3y'' + 3' + y = 0

  • y''' + 3y'' - 3' - y = 0

  • y''' - 3y'' - 3' + y = 0

  • y''' - 3y'' + 3' - y = 0


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363.

The solution of tanydydx = sinx + y + sinx - y is

  • sec(y) = 2cos(x) + c

  • sec(y) = - 2cos(x) + c

  • tan(y) =  - 2cos(x) + c

  • sec2(y) = - 2cos(x) + c


B.

sec(y) = - 2cos(x) + c

tanydydx = sinx + y + sinx - ytanydydx =2sin2x2cos2y2 sinC + sinD = 2sinC + D2cosC - D2 tanydydx = 2sinxcosysinycosydydx = 2sinxcosyOn integrating sinycos2ydy = 2sinxdx  Let t = cosy  dtdy = - siny- dt = sinydy  dtt2 = 2- cosx + c - - 1t = - 2cosx + c     1cos(y) = - 2cos(x) + c       sec(y) = - 2cos(x) + c 


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364.

A family of curves has the differential equation xydydx = 2y2 - x2. Then, the family of curves is

  • y2 = cx2 + x3

  • y2 = cx4 + x3

  • y2 = x + cx4

  • y2 = x2 + cx4


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365.

The solution of the differential equation dydx = yx + ϕyxϕ'yx is

  • yx = k

  • ϕyx = kx

  • yx = k

  • ϕyx = ky


366.

If y = y(x) is the solution of the differential equation 2 + sinxy + 1dydx + cosx = 0 then y(π2)is equal to

  • 13

  • 23

  • 1

  • 43


367.

If u = fr, where r2 = x2 + y2, thenux2 + 2uy2 = ?

  • f''(r)

  • f''(r) +f'(r)

  • f''(r) + 1rf'(r)

  • f''(r) + rf'(r)


368.

If dydx + 2xtanx - y = 1, then sinx - y = ?

  • Ae- x2

  • Ae2x

  • Aex2

  • Ae - 2x


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369.

An integrating factor of the differential equation1 - x2dydx + xy = x41 + x51 - x23 is 

  • 1 - x2

  • x1 - x2

  • x21 - x2

  • 11 - x2


370.

If cos-1yb = 2logx2, where x > 0, thenx2d2ydx2 + xdydx = ?

  • 4y

  • - 4y

  • 0

  • - 8y


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