Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Differential Equations

Multiple Choice Questions

51.

Solution of the differential equation xdy - ydx = 0 represents a

parabola
circle
hyperbola
straight line

52.

The general solution of the differential equation

$\frac{d y}{d x} = e^{y + x} + e^{y - x}$ is

e^{- y} = e^x - e^{- x} + c
e^{- y} = e^{- x} - e^x + c
e^{- y} = e^x + e^-^x + c
e^y = e^x + e^{- x} + c

53.

The order of the differential equation

$\frac{d^{2} y}{d x^{2}} = \sqrt{1 + {(\frac{dy}{dx})}^{2}}$ is

54.

The degree of the differential equation

${[1 + {(\frac{dy}{dx})}^{2}]}^{\frac{5}{3}} = \frac{d^{2} y}{d x^{2}}$ is

1
5
$\frac{10}{3}$
3

55.

The differential equation of all parabolas whose axes are parallel to y-axis, is

$\frac{d^{3} y}{d x^{3}} = 0$
$\frac{d^{2} y}{d x^{2}} = 0$
$\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = 0$
$\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + y = 0$

56.

The solution of the differential equation $\frac{d y}{d x} = e^{y + x} + e^{y - x}$ is

e^{- y} = e^x - e^{- x} + c, c integrating constant
e^{- y} = e^- ^x - e^x + c, c integrating constant
e^{- y} = e^x + e^{- x} + c, c integrating constant
e^{- y} + e^x - e^{- x} = c, c integrating constant

57.
If $x = e^{t} \sin (t), y = e^{t} \cos (t)$ , then $\frac{d^{2} y}{d x^{2}}$ at $x = π$ is

$2 e^{π}$

$\frac{1}{2} e^{π}$

$\frac{1}{2 e^{π}}$

$\frac{2}{e^{π}}$

$\frac{2}{e^{π}}$

$Since, x = e^{t} \sin (t) and y = e^{t} \cos (t) On differentiating w . r . t . t respectively, we get \frac{d x}{d t} = e^{t} \cos (t) + \sin (t) e^{t} and \frac{d y}{d t} = - e^{t} \cos (t) + \sin (t) e^{t} ∴ \frac{d y}{d x} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^{t} (\cos (t) - \sin (t))}{e^{t} (\cos (t) + \sin (t))} = \frac{\cos (t) - \sin (t)}{\cos (t) + \sin (t)}$

$Again differentiating, we get \frac{d^{2} y}{d x^{2}} = \frac{[(\cos (t) + \sin (t)) (- \sin (t) - \cos (t)) - (\cos (t) - \sin (t)) (- \sin (t) + \cos (t))]}{{(\cos (t) + \sin (t))}^{2}} \frac{dt}{dx} = \frac{- {(\sin (t) + \cos (t))}^{2} - {(\cos (t) - \sin (t))}^{2}}{{(\cos (t) + \sin (t))}^{2}} \times \frac{1}{e^{t} (\cos (t) + \sin (t))} = - [\frac{\sin^{2} (t) + \cos^{2} (t) + 2 \sin (t) \cos (t) + \cos^{2} (t) + \sin^{2} (t) - 2 \sin (t) \cos (t)}{e^{t} {(\cos (t) + \sin (t))}^{3}}] = - \frac{2}{e^{t} {(\cos (t) + \sin (t))}^{3}} \Rightarrow {(\frac{d^{2} y}{d x^{2}})}_{(x = π)} = \frac{- 2}{e^{π} {(\cos (π) + \sin (π))}^{3}} = \frac{2}{e^{π}}$