Find the general solution of the differential | vedclass

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

Question

(A) The given differential equation is: $\frac{dy}{dx} = \sin^{-1} x$ Integrating both sides with respect to $x$: $y = \int \sin^{-1} x \, dx$ Using i

Vedclass · Accepted Answer

$y = x \sin^{-1} x + \sqrt{1 - x^2} + C$

Vedclass · Answer

(A) The given differential equation is: $\frac{dy}{dx} = \sin^{-1} x$ Integrating both sides with respect to $x$: $y = \int \sin^{-1} x \, dx$ Using integration by parts,let $u = \sin^{-1} x$ and $dv = dx$. Then $du = \frac{1}{\sqrt{1-x^2}} dx$ and $v = x$. $y = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1-x^2}} dx$ Let $t = 1 - x^2$,then $dt = -2x \, dx$,which implies $x \, dx = -\frac{1}{2} dt$. $y = x \sin^{-1} x - \int \frac{-1/2}{\sqrt{t}} dt$ $y = x \sin^{-1} x + \frac{1}{2} \int t^{-1/2} dt$ $y = x \sin^{-1} x + \frac{1}{2} \cdot \frac{t^{1/2}}{1/2} + C$ $y = x \sin^{-1} x + \sqrt{t} + C$ $y = x \sin^{-1} x + \sqrt{1 - x^2} + C$

Find the general solution of the differential equation: $\frac{dy}{dx} = \sin^{-1} x$

Similar Questions

The solution of the differential equation $e^{-x}(y + 1) dy + (\cos^2 x - \sin 2x)y dx = 0$, given that $y = 1$ when $x = 0$ is

The particular solution of the differential equation $x \, dy + 2y \, dx = 0$, given that $y = 1$ when $x = 2$, is:

If the differential equation $\frac{dy}{dx} + \frac{x}{y} = \frac{a}{y}$,where $a$ is a constant,represents a family of circles,then the radius of the circle is ......

$A$ differentiable function satisfies $3f^2(x) f'(x) = 2x$. Given $f(2) = 1$,then the value of $f(3)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module