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 $\frac{dy}{dx} + 2y = e^{-x}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where

Vedclass · Accepted Answer

$y e^{2x} = e^{x} + c$

Vedclass · Answer

(B) The given differential equation is $\frac{dy}{dx} + 2y = e^{-x}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2$ and $Q = e^{-x}$.The Integrating Factor ($I$.$F$.) is given by $I.F. = e^{\int P dx} = e^{\int 2 dx} = e^{2x}$.The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + c$.Substituting the values,we get $y e^{2x} = \int (e^{-x} \cdot e^{2x}) dx + c$.$y e^{2x} = \int e^{x} dx + c$.$y e^{2x} = e^{x} + c$.

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

Similar Questions

The integrating factor of the differential equation $(1 + t^2) + (x - e^{\tan^{-1} t}) \frac{dt}{dx} = 0$ is

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

The integrating factor of the differential equation $(1+y+x^{2}y)dx + (x+x^{3})dy = 0$ is

The solution of the differential equation $(x + 2y^3) \frac{dy}{dx} - y = 0$ is

Let $y=y(x)$ be the solution of the differential equation $2 \cos x \frac{d y}{d x}=\sin 2 x-4 y \sin x$,where $x \in \left(0, \frac{\pi}{2}\right)$. If $y\left(\frac{\pi}{3}\right)=0$,then $y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module