Let y = y(x) be the solution of the different | vedclass

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

Question

(A) The given differential equation is $x\sqrt{1-x^2} dy + (y\sqrt{1-x^2} - x\cos^{-1}x) dx = 0$.Dividing by $dx$ and $x\sqrt{1-x^2}$, we get $\frac{d

Vedclass · Accepted Answer

$3 - \frac{\pi}{\sqrt{3}}$

Vedclass · Answer

(A) The given differential equation is $x\sqrt{1-x^2} dy + (y\sqrt{1-x^2} - x\cos^{-1}x) dx = 0$.Dividing by $dx$ and $x\sqrt{1-x^2}$, we get $\frac{dy}{dx} + \frac{y}{x} = \frac{\cos^{-1}x}{\sqrt{1-x^2}}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{x}$ and $Q = \frac{\cos^{-1}x}{\sqrt{1-x^2}}$.The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int (1/x) dx} = e^{\ln x} = x$.The solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$, so $yx = \int \frac{x\cos^{-1}x}{\sqrt{1-x^2}} dx$.Let $t = \cos^{-1}x$, then $dt = -\frac{dx}{\sqrt{1-x^2}}$ and $x = \cos t$.Substituting these, $yx = -\int t \cos t dt = -(t \sin t + \cos t) + C = -(\cos^{-1}x \sqrt{1-x^2} + x) + C$.Given $\lim_{x	o 1^-} y(x) = 1$, as $x 	o 1$, $yx 	o 1$. Thus, $1 = -(\cos^{-1}(1) \cdot 0 + 1) + C$, which gives $1 = -1 + C$, so $C = 2$.Therefore, $yx = 2 - x - \sqrt{1-x^2} \cos^{-1}x$.At $x = \frac{1}{2}$, $y(\frac{1}{2}) \cdot \frac{1}{2} = 2 - \frac{1}{2} - \sqrt{1 - (1/2)^2} \cos^{-1}(1/2) = \frac{3}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\pi}{3} = \frac{3}{2} - \frac{\pi}{2\sqrt{3}}$.Multiplying by $2$, we get $y(\frac{1}{2}) = 3 - \frac{\pi}{\sqrt{3}}$.

Let $y = y(x)$ be the solution of the differential equation $x\sqrt{1-x^2} dy + (y\sqrt{1-x^2} - x\cos^{-1}x) dx = 0$, where $x \in (0, 1)$ and $\lim_{x\to 1^-} y(x) = 1$. Then $y\left(\frac{1}{2}\right)$ equals:

Similar Questions

If $y=\sin x+A \cos x$ is the general solution of $\frac{dy}{dx}+f(x)y=\sec x$, then an integrating factor of the differential equation is

If $x=f(y)$ is the solution of the differential equation $(1+y^2)+(x-2 e^{\tan ^{-1} y}) \frac{d y}{d x}=0$,$y \in(-\frac{\pi}{2}, \frac{\pi}{2})$ with $f(0)=1$,then $f(\frac{1}{\sqrt{3}})$ is equal to :

If $\sin x$ is the integrating factor of the linear differential equation $\frac{dy}{dx} + Py = Q$,then $P$ is

The solution of the equation $x\frac{dy}{dx} + 3y = x$ is

Let $y = y(x)$ be the solution curve of the differential equation $x(x^2 + e^x) dy + (e^x(x-2)y - x^3) dx = 0, x > 0$,passing through the point $(1, 0)$. Then $y(2)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module