If y = frac{x}{ln |c x|} (where c is an arbit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $y = \frac{x}{\ln |c x|}$. Taking the reciprocal,we get $\frac{1}{y} = \frac{\ln |c x|}{x}$. This implies $\frac{x}{y} = \ln |c| + \ln |x|$.

Vedclass · Accepted Answer

$-\frac{y^2}{x^2}$

Vedclass · Answer

(D) Given $y = \frac{x}{\ln |c x|}$. Taking the reciprocal,we get $\frac{1}{y} = \frac{\ln |c x|}{x}$. This implies $\frac{x}{y} = \ln |c| + \ln |x|$. Differentiating both sides with respect to $x$: $\frac{d}{dx} \left( \frac{x}{y} \right) = \frac{d}{dx} (\ln |c| + \ln |x|)$. Using the quotient rule on the left side: $\frac{y(1) - x(y')}{y^2} = \frac{1}{x}$. $\frac{y - x \frac{dy}{dx}}{y^2} = \frac{1}{x}$. $y - x \frac{dy}{dx} = \frac{y^2}{x}$. $x \frac{dy}{dx} = y - \frac{y^2}{x}$. $\frac{dy}{dx} = \frac{y}{x} - \frac{y^2}{x^2}$. Comparing this with the given equation $\frac{dy}{dx} = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$,we find that $\phi \left( \frac{x}{y} \right) = -\frac{y^2}{x^2}$.

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	$(s)$ $(0, \frac{\pi}{8})$
	$(t)$ $(-\pi, \pi)$

If $y = \frac{x}{\ln |c x|}$ (where $c$ is an arbitrary constant) is the general solution of the differential equation $\frac{dy}{dx} = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$,then the function $\phi \left( \frac{x}{y} \right)$ is:

Similar Questions

If $2xy^3dx + x^2y^2dy = ydx - xdy$ and $y(2) = 1$,then the value of $y(-1)$ will be (where $y(x)$ denotes the value of $y$ for a given $x$):

The equation of the curve passing through $(3, 4)$ and satisfying the differential equation $y \left( \frac{dy}{dx} \right)^2 + (x - y) \frac{dy}{dx} - x = 0$ can be:

The solution of the differential equation $y dx - x dy + 3x^2 y^2 e^{x^3} dx = 0$ satisfying $y = 1$ when $x = 1$ is:

If the general solution of the differential equation $(y-x+1) dy - (y+x+2) dx = 0$ is $f(x, y, c) = 0$, then the value of $c$ such that $f(1, 1, c) = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module