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 general solution of the differential equation $y y^{\prime} = x \left[ \frac{y^2}{x^2} + \frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi^{\prime}\left(\frac{y^2}{x^2}\right)} \right]$,where $\phi$ is an arbitrary function,is

If $\frac{dy}{dx} = 1 + x + y + xy$ and $y(-1) = 0$,then the function $y$ is

The equation of the curve passing through $\left(\frac{\pi}{6}, 0\right)$ and satisfying the differential equation $(e^y+1) \cos x \, dx + e^y \sin x \, dy = 0$ is:

If the curve $y = f(x)$ passes through the point $(1, e)$ and satisfies the differential equation $dy = y(2 + \log_e x) dx, x > 0$, then $f(e)$ is equal to:

The solution of the differential equation $e^{\frac{dy}{dx}} = x+1$ with the initial condition $y(0) = 5$ for $x \in (-1, \infty)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module