Find the general solution of the differential | vedclass

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

Question

(A) The given differential equation is: $\frac{dy}{dx} + y = 1$ $(y 
eq 1)$.Rearranging the terms,we get:$\frac{dy}{dx} = 1 - y$Separating the variab

Vedclass · Accepted Answer

$y = 1 + Ae^{-x}$

Vedclass · Answer

(A) The given differential equation is: $\frac{dy}{dx} + y = 1$ $(y 
eq 1)$.Rearranging the terms,we get:$\frac{dy}{dx} = 1 - y$Separating the variables,we get:$\frac{dy}{1 - y} = dx$Integrating both sides:$\int \frac{dy}{1 - y} = \int dx$$-\log|1 - y| = x + C_1$$\log|1 - y| = -x - C_1$Taking the exponential of both sides:$|1 - y| = e^{-x - C_1} = e^{-C_1} \cdot e^{-x}$$1 - y = \pm e^{-C_1} e^{-x}$Let $A = \mp e^{-C_1}$,then:$1 - y = Ae^{-x}$$y = 1 - Ae^{-x}$Since $A$ is an arbitrary constant,we can write the general solution as $y = 1 + Ce^{-x}$.

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

Similar Questions

The solution of the differential equation $(1 - x^2)dy + xydx = xy^2dx$ is

Let $f: R \rightarrow R$ be a differentiable function with $f(0)=0$. If $y=f(x)$ satisfies the differential equation $\frac{dy}{dx}=(2+5y)(5y-2)$,then the value of $\lim_{x \rightarrow -\infty} f(x)$ is:

If $\frac{dy}{dx} = \frac{xy + y}{xy + x}$,then the solution of the differential equation is:

The solution of the differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ is

The solution of the differential equation $\frac{dy}{dx} = \frac{xy+y}{xy+x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module