The particular solution of the differential e | vedclass

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

Question

(C) Given the differential equation $\cos \left(\frac{dy}{dx}\right) = a$. Taking $\cos^{-1}$ on both sides,we get $\frac{dy}{dx} = \cos^{-1} a$. Inte

Vedclass · Accepted Answer

$\cos \left(\frac{y-2}{x}\right) = a$

Vedclass · Answer

(C) Given the differential equation $\cos \left(\frac{dy}{dx}\right) = a$. Taking $\cos^{-1}$ on both sides,we get $\frac{dy}{dx} = \cos^{-1} a$. Integrating both sides with respect to $x$,we have $\int dy = \int \cos^{-1} a \, dx$. This gives $y = x \cos^{-1} a + c$ .... $(1)$. Given the initial condition $y(0) = 2$,substitute $x = 0$ and $y = 2$ into equation $(1)$: $2 = 0 \cdot \cos^{-1} a + c \Rightarrow c = 2$. Substituting $c = 2$ back into equation $(1)$,we get $y = x \cos^{-1} a + 2$. Rearranging the terms,$y - 2 = x \cos^{-1} a$,which implies $\frac{y - 2}{x} = \cos^{-1} a$. Taking $\cos$ on both sides,we obtain $\cos \left(\frac{y - 2}{x}\right) = a$.

The particular solution of the differential equation $\cos \left(\frac{dy}{dx}\right) = a$,under the conditions $a \in \mathbb{R}$ and $y(0) = 2$ is

Similar Questions

General solution of the differential equation $(y^3+y)(x^2+1) dy = (xy^4+2y^2x) dx$ is (where $C$ is a constant of integration.)

The solution of the differential equation $\cos(x+y) dy = dx$ given that $y(0) = 0$ is

The general solution of the differential equation $\log \left(\frac{dy}{dx}\right) = ax + by$ is

The solution of the differential equation $\frac{dy}{dx} = 1 - \cos(y-x) \cot(y-x)$ is

Find the general solution of the differential equation: $\frac{dy}{dx} = (1+x^2)(1+y^2)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module