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

Let $f: R \rightarrow R$ be a continuous function satisfying $f(x)=x+\int_0^x f(t) dt$,for all $x \in R$. Then,the number of elements in the set $S=\{x \in R: f(x)=0\}$ is

The integrating factor of $\frac{dy}{dx} + y = \frac{1+y}{x}$ is

Let the slope of the tangent to a curve $y=f(x)$ at $(x, y)$ be given by $2 \tan x(\cos x-y)$. If the curve passes through the point $(\frac{\pi}{4}, 0)$,then the value of $\int_{0}^{\pi / 2} y \, dx$ is equal to

The solution of $\cos y + (x \sin y - 1) \frac{dy}{dx} = 0$ is

If $y(t)$ is a solution of $(1 + t)\frac{dy}{dt} - ty = 1$ and $y(0) = -1$,then $y(1)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module