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Differential Equations

Multiple Choice Questions

141.

The general solution of the differential equation xdy - ydx = y²dx is

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

142.

The order of the differential equation ${(\frac{d^{3} y}{{dx}^{3}})}^{2} + {(\frac{d^{2} y}{{dx}^{2}})}^{2} + {(\frac{dy}{dx})}^{5}$ = 0 is

143.

The solution of dy/dx + ytan(x) = sec(x), y(0) = 0 is

ysec(x) = tan(x)
ytan(x) = sec(x)
tan(x) = ytan(x)
xsec(x) = tan(y)

144.

The differential equation for, which y = a cos(x) + b sin(x) is a solution, is :

$\frac{d^{2} y}{{dx}^{2}} + y = 0$
$\frac{d^{2} y}{{dx}^{2}} - y = 0$
$\frac{d^{2} y}{{dx}^{2}} + (a + b) y = 0$
$\frac{d^{2} y}{{dx}^{2}} = (a + b) y$

145.

The solution of $\frac{dy}{dx} + P (x) y = 0$ is

$y = {ce}^{\int Pdx}$
$y = {ce}^{- \int Pdx}$
$x = {ce}^{- \int Pdy}$
$x = {ce}^{\int Pdy}$

146.

The differential equation of the family of lines passing through the origin is :

$x \frac{dy}{dx} + y = 0$
$x + \frac{dy}{dx} = 0$
$\frac{dy}{dx} = y$
$x \frac{dy}{dx} - y = 0$

147.

The solution of $\frac{dy}{dx} + y = e^{- x}$ ; y(0) = 0 is :

y = e^{- x}(x - 1)
y = xe^{- x}
y = xe^{- x} + 1
y = (x + 1)e^-x

148.

The degree of the differential equation $\frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{3} + 6 y = 0$ is :

149.

The solution of the equation (2y - 1)dx - (2x + 3)dy = 0 is :

$\frac{2 x - 1}{2 y + 3} = c$
$\frac{2 x + 3}{2 y - 1} = c$
$\frac{2 x - 1}{2 y - 1} = c$
$\frac{2 y + 1}{2 x - 3} = c$

150.
Let F denotes the family of ellipses whose centre is at the origin and major axis is the y-axis. Then, equation of the family F is :

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

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

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

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

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

$Equation of family of ellipse is \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 On differentiating w . r . t . x, we get \Rightarrow \frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \frac{dy}{dx} = 0 . . . (i) \Rightarrow \frac{x}{a^{2}} + \frac{y}{b^{2}} \frac{dy}{dx} = 0 . . . (ii) Again differentiating w . r . t . x, we get \frac{1}{a^{2}} + \frac{y}{b^{2}} (\frac{d^{2} y}{{dx}^{2}}) + (\frac{dy}{dx}) = 0 \Rightarrow - \frac{y}{x} \frac{dy}{dx} + y \frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2} = 0 [∵ from (ii)] \Rightarrow xy \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} (x \frac{dy}{dx} - y) = 0$

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Multiple Choice Questions

150. Let F denotes the family of ellipses whose centre is at the origin and major axis is the y-axis. Then, equation of the family F is : d2ydx2 + dydxxdydx - y = 0 xyd2ydx2 + dydxxdydx - y = 0 xyd2ydx2 + dydxxdydx - y = 0 d2ydx2 - dydxxdydx - y = 0