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Differential Equations

Multiple Choice Questions

91.

The solution of the differential equation $\frac{d y}{d x} = \sin (x + y) \tan (x + y) - 1$ is

$\csc (x + y) + \tan (x + y) = x + c$
$x + \csc (x + y) = c$
$x + \tan (x + y) = c$
$x + \sec (x + y) = c$

92.

The differential equation of all straight lines touching the circle x² + y² = a² is

${(y - \frac{dy}{dx})}^{2} = a^{2} [1 + {(\frac{dy}{dx})}^{2}]$
${(y - x \frac{dy}{dx})}^{2} = a^{2} [1 + {(\frac{dy}{dx})}^{2}]$
$(y - x \frac{dy}{dx}) = a^{2} [1 + \frac{dy}{dx}]$
$(y - \frac{dy}{dx}) = a^{2} [1 - \frac{dy}{dx}]$

93.

The differential equation $|\frac{dy}{dx}| + |y| + 3 = 0$ admits

infinite number of solutions
no solutions
a unique solution
many solutions

94.

Solution of the differential equation $xdy - ydx - \sqrt{x^{2} + y^{2}} dx = 0$

$y - \sqrt{x^{2} + y^{2}} = {cx}^{2}$
$y + \sqrt{x^{2} + y^{2}} = {cx}^{2}$
$y + \sqrt{x^{2} + y^{2}} = {cy}^{2}$
$x - \sqrt{x^{2} + y^{2}} = {cy}^{2}$

95.

If $xsin (\frac{x}{y}) dy = [ysin (\frac{y}{x}) - x] dx$ and $y (1) = \frac{π}{2}$ , then the value of $\cos (\frac{y}{x})$ is equal to :

x
$\frac{1}{x}$
$\log (x)$
e^x

96.
The differential equation of the system of all circles of radius r in the xy plane is :

${[1 + {(\frac{dy}{dx})}^{3}]}^{2} = r^{2} {(\frac{d^{2} y}{{dx}^{2}})}^{2}$

${[1 + {(\frac{dy}{dx})}^{3}]}^{2} = r^{2} {(\frac{d^{2} y}{{dx}^{2}})}^{3}$

${[1 + {(\frac{dy}{dx})}^{2}]}^{3} = r^{2} {(\frac{d^{2} y}{{dx}^{2}})}^{2}$

${[1 + {(\frac{dy}{dx})}^{2}]}^{3} = r^{2} {(\frac{d^{2} y}{{dx}^{2}})}^{3}$

${[1 + {(\frac{dy}{dx})}^{2}]}^{3} = r^{2} {(\frac{d^{2} y}{{dx}^{2}})}^{2}$

The equation of the family of circles of radius r is

(x - a)² + (y - b)² = r² ...(i)

where a and b are arbitrary constants.

On differentiating Eq. (i) w.r.t. x, we get

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

$\Rightarrow (x - a) + (y - b) \frac{dy}{dx} = 0$ ...(ii)

On differentiating Eq. (ii) w.r.t. x, we get

$1 + (y - b) \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = 0 \Rightarrow (y - b) = - \frac{1 + {(\frac{dy}{dx})}^{2}}{\frac{d^{2} y}{{dx}^{2}}} . . . (iii)$

On putting the value of (y - b) in Eq. (ii), we get

$(x - a) = \frac{[1 + {(\frac{dy}{dx})}^{2}] \frac{dy}{dx}}{\frac{d^{2} y}{{dx}^{2}}} . . . (iv)$

On putting the value of (y - b) and (x - a), in Eq. (i), we get

$\frac{{[1 + {(\frac{dy}{dx})}^{2}]}^{2} {(\frac{dy}{dx})}^{2}}{{(\frac{d^{2} y}{{dx}^{2}})}^{2}} + \frac{{[1 + {(\frac{dy}{dx})}^{2}]}^{2}}{{(\frac{d^{2} y}{{dx}^{2}})}^{2}} = r^{2} \Rightarrow {[1 + {(\frac{dy}{dx})}^{2}]}^{3} = r^{2} {[\frac{d^{2} y}{{dx}^{2}}]}^{2} This is the required differential equation .$