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Differential Equations

Multiple Choice Questions

551.

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

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

552.

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

553.

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

554.

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

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

555.

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

556.

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

557.

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

558.
Solution of $x^{2} \frac{dy}{dx} - xy = 1 + \cos (\frac{y}{x})$ is

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

$\tan (\frac{y}{x}) = C + \frac{1}{x}$

$\cos (\frac{y}{x}) = 1 + \frac{C}{x}$

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

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

$x^{2} \frac{dy}{dx} - xy = 1 + \cos (\frac{y}{x}) \frac{dy}{dx} - \frac{1}{x} . y = \frac{1}{x^{2}} \cos (\frac{y}{x}) . . . (i) Put y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} ∴ Eq . (i) becomes v = x \frac{dv}{dx} - v = \frac{1}{x^{2}} + \frac{1}{x^{2}} \cos (v) \Rightarrow \frac{dv}{1 + \cos (v)} = \frac{dx}{x^{3}} \Rightarrow \int \sec^{2} (\frac{v}{2}) dv = \frac{- x^{2}}{2} + C \Rightarrow \tan (\frac{v}{2}) = - \frac{1}{2 x^{2}} + C \Rightarrow \tan (\frac{y}{2 x}) = C - \frac{1}{2 x^{2}}$