Integrating factor

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In the case of a linear first order homogeneous ODE such as $au'(x) + bu(x) + c = 0,\,\,u(0) = n\,$ the equation should always be divided by $a$ so the coefficient of the $u'$ term is unity.

$u' + \frac{b}{a}u + \frac{c}{a} = 0\,$

The integrating factor ( $ρ$ ) is found by exponentiating the integral of the coefficient of the $u$ term with respect to $x$ . The integral is indefinate and you don't have to worry about constants of integration.

$\rho = e^{\int \frac{b}{a} dx} = e^{\frac{b}{a}x}\,$

Now, multiply the equation by the integrating factor.

$e^{\frac{b}{a}x}u' + \frac{b}{a}e^{\frac{b}{a}x}u + \frac{c}{a}e^{\frac{b}{a}x} = 0\,$

Notice that the first two terms of the left hand side (LHS) of the equation are the chain rule expansion of

$\frac{d}{dx}\left[e^{\frac{b}{a}x}u\right] = -\frac{c}{a}e^{\frac{b}{a}x}\,$

Integrate both sides with respect to x, and here constants of integration are required again.

$e^{\frac{b}{a}x}u = -\frac{c}{b}e^{\frac{b}{a}x} + c_1\,$

$u(x) = -\frac{c}{b}+c_1e^{-\frac{b}{a}x}\,$

Plugging in 0 for the IC,

$u(0) = -\frac{c}{b} + c_1 = n\,$

$c_1 = \frac{c}{b}+n\,$

So the unique solution for this IC is

$u(x) = -\frac{c}{b} + \left(\frac{c}{b}+n\right)e^{-\frac{b}{a}x}\,$

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