The solution of the differential equation sqr | vedclass

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

Question

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

Vedclass · Accepted Answer

$x = \sin^{-1} y - 1 + c e^{-\sin^{-1} y}$

Vedclass · Answer

(A) Given differential equation is $\sqrt{1-y^2} dx + (x - \sin^{-1} y) dy = 0$. Dividing by $dy$ and $\sqrt{1-y^2}$,we get $\frac{dx}{dy} + \frac{x}{\sqrt{1-y^2}} = \frac{\sin^{-1} y}{\sqrt{1-y^2}}$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{\sqrt{1-y^2}}$ and $Q(y) = \frac{\sin^{-1} y}{\sqrt{1-y^2}}$. The integrating factor is $IF = e^{\int P(y) dy} = e^{\int \frac{1}{\sqrt{1-y^2}} dy} = e^{\sin^{-1} y}$. The solution is $x \cdot IF = \int Q(y) \cdot IF dy + c$. $x e^{\sin^{-1} y} = \int \frac{\sin^{-1} y}{\sqrt{1-y^2}} e^{\sin^{-1} y} dy + c$. Let $t = \sin^{-1} y$,then $dt = \frac{1}{\sqrt{1-y^2}} dy$. $x e^{\sin^{-1} y} = \int t e^t dt + c = (t e^t - e^t) + c = e^t(t - 1) + c$. Substituting $t = \sin^{-1} y$,we get $x e^{\sin^{-1} y} = e^{\sin^{-1} y}(\sin^{-1} y - 1) + c$. Dividing by $e^{\sin^{-1} y}$,we get $x = \sin^{-1} y - 1 + c e^{-\sin^{-1} y}$.

The solution of the differential equation $\sqrt{1-y^2} dx + x dy - \sin^{-1} y dy = 0$ is

Similar Questions

For $x \in R$,let the function $y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 12y = \cos \left(\frac{\pi}{12} x\right)$ with $y(0) = 0$. Then,which of the following statements is/are $TRUE$?

Let $y=y(x)$ be the solution of the differential equation $\cos x(3 \sin x+\cos x+3) dy = (1+y \sin x(3 \sin x+\cos x+3)) dx$; $0 \leq x \leq \frac{\pi}{2}, y(0)=0$. Then,$y\left(\frac{\pi}{3}\right)$ is equal to ..... .

The solution of the differential equation $(1+y^2)+(x-e^{\tan ^{-1} y}) \frac{dy}{dx}=0$ is

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{5}{x(x^5+1)}y=\frac{(x^5+1)^2}{x^7}$,for $x > 0$. If $y(1)=2$,then $y(2)$ is equal to

Find the solution of the differential equation $(e^{y-x}) dy = (e^x - e^y) dx$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module