dydx + y = 1 dydx = 1 y dy( y 1 ) = dx Integrating both sides, we have ∫dy( y 1 ) = ∫ dx log( y 1 ) + C_1 = x + C_2 x + ...

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Solving Linear Differential Equations of First Order

dydx+y=1 y≢1

Question

$\frac{d y}{d x} + y = 1 (y \equiv̸ 1)$

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Solution

$\frac{d y}{d x} + y = 1$
$\frac{d y}{d x} = 1 - y$
$\frac{d y}{(y - 1)} = - d x$
Integrating both sides, we have
$\int \frac{d y}{(y - 1)} = - \int d x$
$log (y - 1) + C_{1} = - x + C_{2}$
$x + log (y - 1) = C_{2} - C_{1}$
$x + log (y - 1) = C (∵ C = C_{2} - C_{1} = constant)$
Therefore, the solution is-
$x + log (y - 1) = C$

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