The particular solution of the differential e | vedclass

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

Question

(A) Given the differential equation: $\cos \left(\frac{dy}{dx}\right) = 0.5$. Taking the inverse cosine on both sides: $\frac{dy}{dx} = \cos^{-1}(0.5)

Vedclass · Accepted Answer

$y = \frac{\pi}{3}x + 1$

Vedclass · Answer

(A) Given the differential equation: $\cos \left(\frac{dy}{dx}\right) = 0.5$. Taking the inverse cosine on both sides: $\frac{dy}{dx} = \cos^{-1}(0.5)$. Since $\cos(60^{\circ}) = 0.5$,we have $\frac{dy}{dx} = \frac{\pi}{3}$. Integrating both sides with respect to $x$: $\int dy = \int \frac{\pi}{3} dx$. This gives $y = \frac{\pi}{3}x + C$. Using the initial condition $y = 1$ at $x = 0$: $1 = \frac{\pi}{3}(0) + C$,which implies $C = 1$. Therefore,the particular solution is $y = \frac{\pi}{3}x + 1$.

The particular solution of the differential equation $\cos \left(\frac{dy}{dx}\right) = 0.5$ with the condition $y = 1$ at $x = 0$ is:

Similar Questions

The solution of the differential equation $\frac{dy}{dx} = (x - y)^2$ when $y(1) = 1$ is:

The solution of the differential equation $x = 1 + xy\frac{dy}{dx} + \frac{(xy)^2}{2!}\left(\frac{dy}{dx}\right)^2 + \frac{(xy)^3}{3!}\left(\frac{dy}{dx}\right)^3 + \dots$ is

The solution of the differential equation $\frac{dy}{dx} = \frac{x-2y+1}{2x-4y}$ is

The solution of the differential equation ${x^2}dy = - 2xydx$ is

The solution of $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module