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  • Find the general solution: dy/dx + y = 1

 3. Find the general solution:

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

Answers (1)

best_answer

Given, in the question

\frac{dy}{dx} + y = 1

\\ \implies \frac{dy}{dx} = 1- y \\ \implies \int\frac{dy}{1-y} = \int dx

(\int\frac{dx}{x} = lnx)

\\ \implies -log(1-y) = x + C\ \ (We\ can\ write\ C= log k) \\ \implies log k(1-y) = -x \\ \implies 1- y = \frac{1}{k}e^{-x} \\

The required general equation

\implies y = 1 -\frac{1}{k}e^{-x}

Posted by

HARSH KANKARIA

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