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Differential Equations

Multiple Choice Questions

201.

The degree and order of the differential equation ${[1 + {(\frac{dy}{dx})}^{3}]}^{\frac{7}{3}} = 7 (\frac{d^{2} y}{{dx}^{2}})$ respectively are

3 and 7
3 and 2
7 and 3
2 and 3

202.

The particular solution of the differential equation

$y (1 + \log (x)) \frac{d x}{d y} - xlog (x) = 0$ , when, x = e, y = e² is

y = exlog(x)
ey = xlog(x)
xy = elog(x)
ylog(x) = ex

203.

If $\sin (x)$ is the integrating factor (IF) of the linear differential equation $\frac{dy}{dx} + Py = Q$ , then P is

$\log (\sin (x))$
$\cos (x)$
$\tan (x)$
$\cot (x)$

204.

The solution of the differential equation

$\frac{dy}{dx} = \tan (\frac{y}{x}) + \frac{y}{x}$ is

$\cos (\frac{y}{x}) = cx$
$\sin (\frac{y}{x}) = cx$
$\cos (\frac{y}{x}) = cy$
$\sin (\frac{y}{x}) = cy$

205.

The differential equation of all parabolas whose axis is Y-axis, is

$x \frac{d^{2} y}{{dx}^{2}} - \frac{dy}{dx} = 0$
$x \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} = 0$
$\frac{d^{2} y}{{dx}^{2}} - y = 0$
$\frac{d^{2} y}{{dx}^{2}} - \frac{dy}{dx} = 0$

206.

The particular solution of the differential equation xdy + 2ydx = 0, when x = 2, y = 1 is

xy = 4
x²y = 4
xy² = 4
x²y² = 4

207.
The general solution of the equation $\frac{dy}{dx} = \frac{y^{2} - x}{2 y (x + 1)}$ is

y² = (1 + x)log(1 + x) - c

$y^{2} = (1 + x) \log (\frac{c}{(1 - x)}) - 1$

$y^{2} = (1 - x) \log (\frac{c}{(1 - x)}) - 1$

$y^{2} = (1 + x) \log (\frac{c}{(1 + x)}) - 1$

$y^{2} = (1 + x) \log (\frac{c}{(1 + x)}) - 1$

$\frac{dy}{dx} = \frac{y^{2} - x}{2 y (x + 1)} . . . (i) \Rightarrow y \frac{dy}{dx} = \frac{y^{2} - x}{2 (x + 1)} \Rightarrow y \frac{dy}{dx} - \frac{y^{2}}{2 (x + 1)} = - \frac{x}{2 (x + 1)} . . . (ii) Put y^{2} = t By differentiating both side w . r . t .' x', we get 2 y \frac{dy}{dx} = \frac{dt}{dx} ∴ Eq . (ii) reduces to \frac{1}{2} \frac{dt}{dx} - \frac{t}{2 (x + 1)} = \frac{- x}{2 (x + 1)} This is linear differential equation with P = \frac{- 1}{x + 1} and Q = \frac{- x}{x + 1}$

$∴ IF = e^{\int Pdx} = e^{\int \frac{- 1}{(x + 1)} dx} = e^{- \log (x + 1)} = e^{\log (\frac{1}{x + 1})} = \frac{1}{x + 1} ∴ Required solution will be t IF = \int Q (IF) dx + \log (c) [Here, \log (c) is a constant] y^{2} . \frac{1}{1 + x} = \int (\frac{- x}{x + 1} \times \frac{1}{x + 1}) dx + \log (c) \Rightarrow \frac{y^{2}}{1 + x} = - \int \frac{(x + 1) - 1}{{(x + 1)}^{2}} dx + \log (c) \Rightarrow \frac{y^{2}}{1 + x} = - [\int \frac{1}{1 + x} dx - \int \frac{1}{{(1 + x)}^{2}} dx] + \log (c) \Rightarrow \frac{y^{2}}{1 + x} = - [\log (1 + x) + \frac{1}{1 + x}] + \log (c) \Rightarrow \frac{y^{2}}{1 + x} = \log (\frac{c}{1 + x}) - \frac{1}{1 + x} \Rightarrow y^{2} = (1 + x) \log (\frac{c}{1 + x}) - 1$

208.

The general solution of the differential equation $\frac{dy}{dx} + \sin (\frac{x + y}{2}) = \sin (\frac{x - y}{2})$ is

$\log_{e} |\tan (\frac{y}{2})| = - 2 \sin (\frac{x}{2}) + C$
$\log_{e} |\tan (\frac{y}{4})| = 2 \sin (\frac{x}{2}) + C$
$\log_{e} (\tan (\frac{y}{2})) = - 2 \sin (\frac{x}{2}) + C$
$\log_{e} (\tan (\frac{y}{4})) = - 2 \sin (\frac{x}{2}) + C$

209.

The function y specified implicitly by the relation $\int_{0}^{y} e^{t} dt + \int_{0}^{x} \cos (t) dt$ = 0 satisfies the differential equation

$e^{2 y} (\frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2}) = \sin (x)$
$e^{y} (\frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2}) = \sin (2 x)$
$e^{y} (2 \frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2}) = \sin (x)$
$e^{y} (\frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2}) = \sin (x)$

210.

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

$e^{y} = 2 e^{x} + \frac{x^{3}}{3} + C$
$e^{- y} = 2 e^{x} + \frac{x^{- 3}}{3} + C$
$e^{- y} = 2 e^{x} + \frac{x^{3}}{3} + C$
$e^{y} = 2 e^{- x} + \frac{x^{3}}{3} + C$

Previous Year Papers

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Multiple Choice Questions

207. The general solution of the equation dydx = y2 - x2yx + 1 is y2 = (1 + x)log(1 + x) - c y2 = 1 + xlogc1 - x - 1 y2 = 1 - xlogc1 - x - 1 y2 = 1 + xlogc1 + x - 1

207.
The general solution of the equation $\frac{dy}{dx} = \frac{y^{2} - x}{2 y (x + 1)}$ is

y² = (1 + x)log(1 + x) - c

$y^{2} = (1 + x) \log (\frac{c}{(1 - x)}) - 1$

$y^{2} = (1 - x) \log (\frac{c}{(1 - x)}) - 1$

$y^{2} = (1 + x) \log (\frac{c}{(1 + x)}) - 1$