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211.
The solution of the differential equation $(x + 2 y^{3}) \frac{dy}{dx} = Y$ is

y³ + Cx = y

$\frac{{xy}^{4}}{2} + xy = Cy$

y³ + Cy = x

x + 2y³ = y + C

y³ + Cy = x

$Given, (x + 2 y^{3}) \frac{dy}{dx} = y y \frac{dx}{dy} = x + 2 y^{3} \Rightarrow \frac{dx}{dy} - \frac{1}{y} x = 2 y^{2} This is the form of \frac{dx}{dy} + Px = Q where, P = - \frac{1}{y} and Q = 2 y^{2} Thus, the given equation is linear .$

$∴ IF = e^{\int Pdy} = e^{\int - \frac{1}{y} dy} = e^{- \log (y)} = e^{\log {(y)}^{- 1}} = y^{- 1} = \frac{1}{y}$

$So, the required solution is x . IF = \int (Q . IF) dy + C \Rightarrow x . \frac{1}{y} = \int (2 y^{2} . \frac{1}{y}) dy + C \Rightarrow x . \frac{1}{y} = \int 2 ydy + C = y^{2} + C \Rightarrow x = y^{3} + Cy$

212.

The solution ofthe differential equation

$\frac{dy}{dx} = \frac{y (\log (y) - \log (x) + 1)}{x}$ is

x = ye^cy
y = xe^cy
x = ye^cx
None of these

213.

The solution of the differential equation $\frac{dy}{dx} + \sin (\frac{y + x}{2}) + \sin (\frac{y - x}{2}) = 0$ is

$\log (\tan (\frac{y}{2})) = C - 2 \sin (x)$
$\log (\tan (\frac{y}{4})) = C - 2 \sin (\frac{x}{2})$
$\log (\tan (\frac{y}{2} + \frac{π}{4})) = C - 2 \sin (x)$
$\log (\tan (\frac{y}{2} + \frac{π}{4})) = C - 2 \sin (\frac{x}{2})$

214.

The solution of differential equation $(x^{2} + y^{2}) - 2 xy \frac{dy}{dx} = 0$ is

$x^{2} + y^{2} = xC$
$x^{2} - y^{2} = xC$
x² + y² = C
x² - y² = C

215.

The solution of differential equation $\frac{dy}{dx} = \frac{x (2 \log (x) + 1)}{\sin (y) + ycos (y)}$ is

$ysin (y) = x^{2} \log (x) + C$
$y = x^{2} + \log (x) + C$
$ysin (y) = x^{2} + C$
None of these

216.

Solution of $2 ysin (x) \frac{dy}{dx} = 2 \sin (x) \cos (x) - y^{2} \cos (x), x = \frac{π}{2}$ , y = 1 is given by

y² = sin(x)
y = sin²(x)
y² = cos(x) + 1
None of these

217.

Solution of $x^{2} \frac{dy}{dx} - xy = 1 + \cos (\frac{y}{x})$ is

$\tan (\frac{y}{2 x}) = C - \frac{1}{2 x^{2}}$
$\tan (\frac{y}{x}) = C + \frac{1}{x}$
$\cos (\frac{y}{x}) = 1 + \frac{C}{x}$
$x^{2} = (C + x^{2}) \tan (\frac{y}{x})$

218.

The solution of $\frac{dy}{dx} = \cos (x) (2 - ycsc (x))$ where y = 2, when x = $\frac{π}{2}$ is

$y = \sin (x) + \csc (x)$
$y = \tan (\frac{x}{2}) + \cot (\frac{x}{2})$
$y = \frac{1}{\sqrt{2}} \sec (\frac{x}{2}) + \sqrt{2} \cos (\frac{x}{2})$
None of the above

219.

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

$\tan (x + y) + \sec (x + y) = x + C$
$\tan (x + y) - \sec (x + y) = x + C$
$\tan (x + y) - \sec (x + y) + x + C = 0$
None of the above

220.

The solution of differential equation

$4 xy \frac{dy}{dx} = \frac{3 {(1 + x)}^{2} (1 + y^{2})}{(1 + x^{2})}$ is

$\log (1 + y) = \log (x) + 2 \tan (x) + C$
$\log (1 + y^{2}) = 3 \log (\frac{1}{x}) + 6 \tan^{- 1} (x) + C$
$\log (1 + y^{2}) = 3 \log (x) + 6 \tan^{- 1} (x) + C$
None of the above

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Differential Equations

Multiple Choice Questions

211.
The solution of the differential equation $(x + 2 y^{3}) \frac{dy}{dx} = Y$ is

y³ + Cx = y

$\frac{{xy}^{4}}{2} + xy = Cy$

y³ + Cy = x

x + 2y³ = y + C

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

211. The solution of the differential equation x + 2y3dydx = Y is y3 + Cx = y xy42 + xy = Cy y3 + Cy = x x + 2y3 = y + C

211.
The solution of the differential equation $(x + 2 y^{3}) \frac{dy}{dx} = Y$ is

y³ + Cx = y

$\frac{{xy}^{4}}{2} + xy = Cy$

y³ + Cy = x

x + 2y³ = y + C