The solution of dydx + y = e- x; y(

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 Multiple Choice QuestionsMultiple Choice Questions

481.

The solution of the differential equation xdydx - yx2 - y2 = 10x2 is

  • sin-1yx - 5x2 = C

  • sin-1yx = 10x2 + C

  • yx = 5x2 + C

  • sin-1yx = 10x2 + Cx


482.

The general solution of the differential equation xdy - ydx = y2dx is

  • y = xC - x

  • x = 2yC + x

  • y = C + x2x

  • y = 2xC + x


483.

The order of the differential equation d3ydx32 + d2ydx22 + dydx5 = 0 is

  • 3

  • 4

  • 1

  • 5


484.

The solution of dy/dx + ytan(x) = sec(x), y(0) = 0 is

  • ysec(x) = tan(x)

  • ytan(x) = sec(x)

  • tan(x) = ytan(x)

  • xsec(x) = tan(y)


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485.

The differential equation for, which y = a cos(x) + b sin(x) is a solution, is :

  • d2ydx2 + y = 0

  • d2ydx2 - y = 0

  • d2ydx2 + a + by = 0

  • d2ydx2 = a + by


486.

The solution of dydx + Pxy = 0 is

  • y = cePdx

  • y = ce- Pdx

  • x = ce- Pdy

  • x = cePdy


487.

The differential equation of the family of lines passing through the origin is :

  • xdydx + y = 0

  • x + dydx = 0

  • dydx = y

  • xdydx - y = 0


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488.

The solution of dydx + y = e- x; y(0) = 0 is :

  • y = e- x(x - 1)

  • y = xe- x

  • y = xe- x + 1

  • y = (x + 1)e-x


B.

y = xe- x

The given differential equation is

dydx + y = e- xOn comparing with dydx + Py = Q, we get         P = 1 and Q = e- x    IF = ePdx = ex Required solution is      yex = ex . e- x dx + c yex = x + cAt y = 0 and x = 0    c = 0,     y = xe- x


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489.

The degree of the differential equation d2ydx2 + dydx3 + 6y = 0 is :

  • 1

  • 3

  • 2

  • 5


490.

The solution of the equation (2y - 1)dx - (2x + 3)dy = 0 is  :

  • 2x - 12y + 3 = c

  • 2x + 32y - 1 = c

  • 2x - 12y - 1 = c

  • 2y + 12x - 3 = c


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