The differential equation frac{dy}{dx} = frac | vedclass

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

Question

(D) Given the differential equation: $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$ Separate the variables: $\int \frac{y}{\sqrt{1-y^2}} dy = \int dx$ Let $

Vedclass · Accepted Answer

fixed radius of $1$ unit and variable centre along the $X$-axis.

Vedclass · Answer

(D) Given the differential equation: $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$ Separate the variables: $\int \frac{y}{\sqrt{1-y^2}} dy = \int dx$ Let $u = 1-y^2$,then $du = -2y dy$,so $y dy = -\frac{1}{2} du$. The integral becomes: $-\frac{1}{2} \int u^{-1/2} du = x + C$ $-\frac{1}{2} (2u^{1/2}) = x + C$ $-\sqrt{1-y^2} = x + C$ Square both sides: $1-y^2 = (x+C)^2$ $(x+C)^2 + y^2 = 1$ This is the equation of a circle $(x-h)^2 + (y-k)^2 = r^2$ where $h = -C$,$k = 0$,and $r = 1$. Thus,the radius is fixed at $1$ unit and the centre $(-C, 0)$ varies along the $X$-axis.

The differential equation $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{y}$ determines a family of circles with

Similar Questions

If the population grows at the rate of $8 \%$ per year,then the time taken for the population to be doubled is $\quad$ (given $\log 2=0.6912$ ) (in $years$)

The family of curves whose $x$ and $y$ intercepts of a tangent at any point are respectively double the $x$ and $y$ coordinates of that point is

The population $p(t)$ at time $t$ of a certain mouse species satisfies the differential equation $\frac{dp(t)}{dt} = 0.5p(t) - 450$. If $p(0) = 850$,then the time at which the population becomes zero is:

The equation of the curve passing through $\left(2, \frac{9}{2}\right)$ and having the slope $\left(1-\frac{1}{x^2}\right)$ at $(x, y)$ is

If it is mentioned that for a curve passing through $(3,4)$,the slope of the curve at any point is the reciprocal of twice the ordinate of that point,then that curve is a

Vedclass Test Series

Exam Paper Generator

Online Exam Module