The solution of differential equation xcos2ydx = y

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Differential Equations

Multiple Choice Questions

221.

The order and power of differential equation

$\frac{d^{2} y}{{dx}^{2}} + 7 \frac{dy}{dx} + \int ydx = \sin (x)$ is

1, 3
3, 1
1, 2
2, 1

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222.
The solution of differential equation ${xcos}^{2} (y) dx = {ycos}^{2} (x) dx$ is

$xtan (x) - ytan (y) - \log (\sec (x) / \sec (y)) = c$

$ytan (x) - xtan (x) - \log (\sec (x) . \sec (y)) = c$

$xtan (x) - ytan (y) + \log (\sec (x) . \sec (y)) = c$

None of the above

A.

$xtan (x) - ytan (y) - \log (\sec (x) / \sec (y)) = c$

${xcos}^{2} (y) dx = {ycos}^{2} (x) dx \Rightarrow \frac{x}{\cos^{2} (x)} dx = \frac{y}{\cos^{2} (y)} dy \Rightarrow {xsec}^{2} (x) dx = {ysec}^{2} (y) dy On integration, we get \Rightarrow xtan (x) - \int 1 \tan (x) dx = y \tan (y) - \int 1 . \tan (y) dy \Rightarrow xtan (x) - \log (\sec (x)) = ytan (y) - \log (\sec (y)) + c \Rightarrow xtan (x) - ytan (y) - (\log (\sec (x)) - \log (\sec (y))) = c \Rightarrow xtan (x) - ytan (y) - \log (\sec (x) / \sec (y)) = c$

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223.

Differential equation of those circles which passes through origin and their centres lie on y-axis will be

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

224.

The differential equation of all circles touching the axis of y at origin and centre on the x-axis is given by

$xy \frac{dy}{dx} - x^{2} + y^{2} = 0$
$2 xy \frac{dy}{dx} - x^{2} - y^{2} = 0$
$(x^{2} + y^{2}) \frac{dy}{dx} - 2 xy = 0$
None of these

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225.

The solution of the differential equation

$(e^{- 2 \sqrt{x}} - \frac{y}{\sqrt{x}}) \frac{dx}{dy} = 1$ is given by

$e^{2 \sqrt{x}} = 2 \sqrt{x} + c$
${ye}^{- 2 \sqrt{x}} = \sqrt{x} + c$
$y = \sqrt{x}$
None of these

226.

The solution of the differential equation

$\frac{dy}{dx} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$ is

$\sin^{- 1} (y) - \sin^{- 1} (x) = c$
$\sin^{- 1} (y) + \sin^{- 1} (x) = c$
$\sin^{- 1} (xy) = 2$
None of these

227.

The solution of differential equation (1 + x)ydx + (1 - y)x dy = 0 is

$\log_{e} (xy) + x - y = c$
$\log_{e} (\frac{x}{y}) + x + y = c$
$\log_{e} (\frac{x}{y}) - x + y = c$
$\log_{e} (xy) - x + y = c$

228.

The differential equation of all circles which passes through the origin and whose centre lies on y-axis is

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

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229.

The general solution of y²dx + (x² - xy + y²)dy = 0 is

$\tan^{- 1} (\frac{x}{y}) + \log (y) + c = 0$
$2 \tan^{- 1} (\frac{x}{y}) + \log (x) + c = 0$
$\log (y + \sqrt{x^{2} + y^{2}}) + \log (y) + c = 0$
$\sinh^{- 1} (\frac{x}{y}) + \log (y) + c = 0$

230.

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

$y = \frac{1}{4} e^{- 2 x} + \frac{cx}{2} + d$
$y = \frac{1}{4} e^{- 2 x} + cx + d$
$y = \frac{1}{4} e^{- 2 x} + {cx}^{2} + d$
$y = \frac{1}{4} e^{- 2 x} + {cx}^{3} + d$

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