The solution of the differential equation fra | vedclass

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

Question

(B) The given differential equation is $\frac{dy}{dx} - 2y 	an 2x = e^x \sec 2x$. This is a linear differential equation of the form $\frac{dy}{dx} +

Vedclass · Accepted Answer

$y \cos 2x = e^x + C$

Vedclass · Answer

(B) The given differential equation is $\frac{dy}{dx} - 2y 	an 2x = e^x \sec 2x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -2 	an 2x$ and $Q = e^x \sec 2x$. First,we find the integrating factor $(IF)$: $IF = e^{\int P dx} = e^{\int -2 	an 2x dx} = e^{-\ln(\sec 2x)} = e^{\ln(\cos 2x)} = \cos 2x$. The general solution is given by $y \cdot IF = \int (Q \cdot IF) dx + C$. Substituting the values: $y \cos 2x = \int (e^x \sec 2x \cdot \cos 2x) dx + C$. Since $\sec 2x \cdot \cos 2x = 1$,we have: $y \cos 2x = \int e^x dx + C$. $y \cos 2x = e^x + C$,where $C$ is the constant of integration.

The solution of the differential equation $\frac{dy}{dx} - 2y \tan 2x = e^x \sec 2x$ is

Similar Questions

If $\int\limits_a^x {t\,y(t)dt} = x^2 + y(x)$,then $y$ as a function of $x$ is

The integrating factor of $y + \frac{d}{dx}(xy) = x(\sin x + \log x)$ is

If $u(x)$ and $v(x)$ are two independent solutions of the differential equation $\frac{d^{2} y}{d x^{2}}+b \frac{d y}{d x}+c y=0$, then which of the following is also a solution of the given differential equation?

The solution of the differential equation $\frac{dy}{dx} + \frac{y}{2} \sec x = \frac{\tan x}{2y}$,where $0 \le x < \frac{\pi}{2}$,and $y(0) = 1$,is given by

The integrating factor of the differential equation $y dx - (x + 2y^2) dy = 0$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module