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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: $\frac{dy}{dx} + y = 1$ $(y \neq 1)$.

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