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JEE Mathematics Solved Question Paper 2006

Multiple Choice Questions

31.

The solution of the differential equation $\frac{dy}{dx} + \frac{2 yx}{1 + x^{2}} = \frac{1}{{(1 + x^{2})}^{2}}$ is :

y(1 + x²) = c + tan^-1(x)
$ylog (1 + x^{2}) = c + \tan^{- 1} (x)$
$\frac{y}{1 + x^{2}} = c + \tan^{- 1} (x)$
$y (1 + x^{2}) = c + \sin^{- 1} (x)$

32.

The solution of the differential equation $xdy - ydx = \sqrt{x^{2} + y^{2}} dx$ is :

$x + \sqrt{x^{2} + y^{2}} = {cx}^{2}$
$y - \sqrt{x^{2} + y^{2}} = cx$
$x - \sqrt{x^{2} + y^{2}} = cx$
$y + \sqrt{x^{2} + y^{2}} = {cx}^{2}$

33.
The solution of the differential equation $\frac{dy}{dx} = e^{x - y} + x^{2} e^{- y}$ is :

y = e^{x - y} + x²e^{- y} + c

$e^{y} - e^{x} = \frac{1}{3} x^{3} + c$

$e^{y} + e^{x} = \frac{1}{3} x^{3} + c$

$e^{x} - e^{y} = \frac{1}{3} x^{3} + c$

$e^{y} - e^{x} = \frac{1}{3} x^{3} + c$

$Given that, \frac{dy}{dx} = e^{x - y} + x^{2} e^{- y} \Rightarrow \frac{dy}{dx} e^{- y} (e^{x} + x^{2}) \Rightarrow \int e^{y} dy = \int e^{x} dx + \int x^{2} dx \Rightarrow e^{y} = e^{x} + \frac{1}{3} x^{3} + c \Rightarrow e^{y} - e^{x} = \frac{1}{3} x^{3} + c$