Verify that the given function $y - \cos y = x$ is a solution of the differential equation $(y \sin y + \cos y + x) y' = y$.
(N/A) Given equation: $y - \cos y = x$ ..........$(1)$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y) - \frac{d}{dx}(\cos y) = \frac{d}{dx}(x)$
$y' - (-\sin y) y' = 1$
$y'(1 + \sin y) = 1$
$y' = \frac{1}{1 + \sin y}$ ..........$(2)$
Now,consider the $L.H.S.$ of the differential equation:
$L.H.S. = (y \sin y + \cos y + x) y'$
Substitute $x = y - \cos y$ from equation $(1)$:
$L.H.S. = (y \sin y + \cos y + y - \cos y) y'$
$L.H.S. = (y \sin y + y) y'$
$L.H.S. = y(1 + \sin y) y'$
Substitute $y'$ from equation $(2)$:
$L.H.S. = y(1 + \sin y) \cdot \frac{1}{1 + \sin y}$
$L.H.S. = y = R.H.S.$
Since $L.H.S. = R.H.S.$,the given function is indeed a solution of the differential equation.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y) - \frac{d}{dx}(\cos y) = \frac{d}{dx}(x)$
$y' - (-\sin y) y' = 1$
$y'(1 + \sin y) = 1$
$y' = \frac{1}{1 + \sin y}$ ..........$(2)$
Now,consider the $L.H.S.$ of the differential equation:
$L.H.S. = (y \sin y + \cos y + x) y'$
Substitute $x = y - \cos y$ from equation $(1)$:
$L.H.S. = (y \sin y + \cos y + y - \cos y) y'$
$L.H.S. = (y \sin y + y) y'$
$L.H.S. = y(1 + \sin y) y'$
Substitute $y'$ from equation $(2)$:
$L.H.S. = y(1 + \sin y) \cdot \frac{1}{1 + \sin y}$
$L.H.S. = y = R.H.S.$
Since $L.H.S. = R.H.S.$,the given function is indeed a solution of the differential equation.
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Match the statements/expressions in Column $I$ with the open intervals in Column $II$.
Column $I$ Column $II$ $(A)$ Interval contained in the domain of definition of non-zero solutions of the differential equation $(x-3)^2 y^{\prime}+y=0$ $(p)$ $(-\frac{\pi}{2}, \frac{\pi}{2})$ $(B)$ Interval containing the value of the integral $\int_1^5(x-1)(x-2)(x-3)(x-4)(x-5) dx$ $(q)$ $(0, \frac{\pi}{2})$ $(C)$ Interval in which at least one of the points of local maximum of $\cos^2 x+\sin x$ lies $(r)$ $(\frac{\pi}{8}, \frac{5\pi}{4})$ $(D)$ Interval in which $\tan^{-1}(\sin x+\cos x)$ is increasing $(s)$ $(0, \frac{\pi}{8})$ $(t)$ $(-\pi, \pi)$
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| $(A)$ Interval contained in the domain of definition of non-zero solutions of the differential equation $(x-3)^2 y^{\prime}+y=0$ | $(p)$ $(-\frac{\pi}{2}, \frac{\pi}{2})$ |
| $(B)$ Interval containing the value of the integral $\int_1^5(x-1)(x-2)(x-3)(x-4)(x-5) dx$ | $(q)$ $(0, \frac{\pi}{2})$ |
| $(C)$ Interval in which at least one of the points of local maximum of $\cos^2 x+\sin x$ lies | $(r)$ $(\frac{\pi}{8}, \frac{5\pi}{4})$ |
| $(D)$ Interval in which $\tan^{-1}(\sin x+\cos x)$ is increasing | $(s)$ $(0, \frac{\pi}{8})$ |
| $(t)$ $(-\pi, \pi)$ |
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