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Differential Equations

Multiple Choice Questions

541.

The differential equation of the family of circles touching Y-axis at the origin is

$(x^{2} + y^{2}) \frac{dy}{dx} - 2 xy = 0$
$(x^{2} - y^{2}) + 2 xy \frac{dy}{dx} = 0$
$(x^{2} - y^{2}) \frac{dy}{dx} - 2 xy = 0$
$(x^{2} + y^{2}) \frac{dy}{dx} + 2 xy = 0$

542.

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

543.
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

y = exlog(x)

$Given, differential equation is y (1 + \log (x)) \frac{d x}{d y} - xlog (x) = 0 \Rightarrow \frac{(1 + \log (x)) dx}{xlog (x)} = \frac{dy}{y} \Rightarrow (\frac{1}{xlog (x)} + \frac{1}{x}) dx = \frac{1}{y} dy On integrating both sides, we get$

$\int (\frac{1}{xlog (x)} + \frac{1}{x}) dx = \int \frac{1}{y} dy Put \log (x) = t \Rightarrow \frac{1}{x} dx = dt ∴ \int \frac{1}{t} dt + \int \frac{1}{x} dx = \int \frac{1}{y} dy \Rightarrow \log (t) + \log (x) = \log (y) + \log (c) \Rightarrow \log (tx) = \log (yc) \Rightarrow tx = yc \Rightarrow xlog (x) = yc When x = e and y = e^{2} ∴ elog (e) = e^{2} c \Rightarrow e \times 1 = e^{2} c \Rightarrow c = \frac{1}{e} ∴ xlog (x) = \frac{y}{e} \Rightarrow y = exlog (x)$

544.

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

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

545.

The solution of the differential equation

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

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

546.

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

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

547.

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

xy = 4
x²y = 4
xy² = 4
x²y² = 4

548.

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

549.

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

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

550.

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

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

Previous Year Papers

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Multiple Choice Questions

543. The particular solution of the differential equationy1 + logxdxdy - xlogx = 0, when, x = e, y = e2 is y = exlog(x) ey = xlog(x) xy = elog(x) ylog(x) = ex

543.
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