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Question

(A) 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,

Vedclass · Accepted Answer

$\log \left|\frac{1}{1-y}\right| = x + C$

Vedclass · Answer

(A) 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 have $\frac{dy}{1-y} = dx$.Integrating both sides,we get $\int \frac{dy}{1-y} = \int dx$.This gives $-\log |1-y| = x + C$.Using the property of logarithms,$-\log |a| = \log |\frac{1}{a}|$,we get $\log \left|\frac{1}{1-y}ight| = x + C$.

General solution of differential equation $\frac{dy}{dx} + y = 1$ $(y \neq 1)$ is

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