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Differential Equations

Multiple Choice Questions

351.

$The solution of (1 + x^{2}) \frac{dy}{dx} + 2 xy - 4 x^{2} = 0 is :$

$3 x (1 + y^{2}) = 4 y^{3} + c$
3y(1 + x²) = 4x³ + c
3x(1 + y²) = 4y³ + c
3y(1 + y²) = 4x³ + c

352.
$The solution of \frac{dx}{dy} + \frac{x}{y} = x^{2} is :$

$\frac{1}{y} = cx - xlog (x)$

$\frac{1}{x} = cy - ylog (y)$

$\frac{1}{x} = cx + xlog (y)$

$\frac{1}{y} = cx - ylog (x)$

$\frac{1}{x} = cy - ylog (y)$

$\frac{dx}{dy} + \frac{x}{y} = x^{2} \Rightarrow \frac{1}{x^{2}} \frac{dx}{dy} + \frac{1}{xy} = 1 Put \frac{1}{x} = t \Rightarrow - \frac{1}{x^{2}} \frac{dx}{dy} = \frac{dt}{dy} ∴ - \frac{dt}{dy} + \frac{t}{y} = 1 \Rightarrow \frac{dt}{dy} - \frac{t}{y} = - 1 On compairing with \frac{dy}{dx} + Py = Q, we get P = - \frac{1}{y}, Q = - 1 I . F . = e^{\int pdx} = e^{- \int \frac{1}{y} dy} = \frac{1}{y} The solution is t (I . F .) = \int Q (I . F .) dy + c t . \frac{1}{y} = \int (- 1) . \frac{1}{y} dy + c \frac{1}{x} \frac{1}{y} = - \log (y) + c \frac{1}{x} = cy - y \log (y)$

353.

The differential equation obtained by eliminating the arbitrary constants a and b from xy = ae^x + be^{- x} is

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

354.

$The solution of (x + y + 1) \frac{dy}{dx} = 1 is$

y = (x + 2) + ce^x
y = - (x + 2) + ce^x
x = - (y + 2) + ce^y
x = (y + 2)² + ce^y

355.

$The solution of \frac{dy}{dx} = \frac{y^{2}}{xy - x^{2}} is$

$e^{\frac{y}{x}} = kx$
$e^{\frac{y}{x}} = ky$
$e^{\frac{x}{y}} = kx$
$e^{\frac{- y}{x}} = ky$

356.

$The solution of \frac{dy}{dx} + 1 = e^{x + y} is$

$e^{- (x + y)} + x + c = 0$
$e^{- (x + y)} - x + c = 0$
$e^{(x + y)} + x + c = 0$
$e^{(x + y)} - x + c = 0$

357.

The solution of the differential equation

$\frac{dy}{dx} = \frac{xy + y}{xy + x} is$

$x + y = \log (\frac{cy}{x})$
$x + y = \log (cxy)$
$x - y - \log (\frac{cx}{y})$
$y - x = \log (\frac{cx}{y})$

358.

The solution of the differential equation

$\frac{dy}{dx} = \frac{x - 2 y + 1}{2 x - 4 y} is$

(x - 2y)² + 2x = c
(x - 2y)² + x = c
(x - 2y)² + 2x² = c
(x - 2y) + x² = c

359.

The solution of the differential equation $\frac{dy}{dx} - ytan (x) = e^{x} \sec (x) is$

$y = e^{x} \cos (x) + c$
$ycos (x) = e^{x} + c$
$y = e^{x} \sin (x) + c$
$ysin (x) = e^{x} + c$

360.

The solution of the differential equation

${xy}^{2} dy - (x^{3} + y^{3}) dx = 0 is$

$y^{3} = 3 x^{3} + c$
$y^{3} = 3 x^{3} \log (cx)$
$y^{3} = 3 x^{3} + \log (cx)$
$y^{3} + 3 x^{3} = \log (cx)$

Previous Year Papers

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Multiple Choice Questions

352. The solution of dxdy + xy = x2 is : 1y = cx - xlogx 1x = cy - ylogy 1x = cx + xlogy 1y = cx - ylogx

352.
$The solution of \frac{dx}{dy} + \frac{x}{y} = x^{2} is :$

$\frac{1}{y} = cx - xlog (x)$

$\frac{1}{x} = cy - ylog (y)$

$\frac{1}{x} = cx + xlog (y)$

$\frac{1}{y} = cx - ylog (x)$