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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.

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

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

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

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

$Given (x + y + 1) \frac{dy}{dx} = 1 \Rightarrow \frac{dx}{dy} = x + y + 1 \Rightarrow \frac{dx}{dy} - x = y + 1 which is linear ∴ IF = e^{\int - 1 dy} = e^{- y} \Rightarrow {xe}^{- y} = \int (y + 1) e^{- y} dy + c \Rightarrow {xe}^{- y} = - {ye}^{- y} + \int 1 . e^{- y} dy + e^{- y} . (- 1) + c \Rightarrow {xe}^{- y} = - {ye}^{- y} - e^{- y} - e^{- y} + c \Rightarrow {xe}^{- y} = - (y + 2) e^{- y} + c \Rightarrow x = - (y + 2) + {ce}^{y}$