Solving First Order Linear ODEs
Most first order Linear ⚠ ODEs can be simplified to the following form-
{$ u’ + p(t)u = g(t)$}, where {$u’ = f(t,u)$}
Keep in mind that our unknown u is a function, not a variable.
Simple Cases
The simplest cases are when the coefficient functions are simply constants- we’ll define {$p(t) = \alpha$} and {$g(t) = \beta$}. In this case, we have a simple explicit solution-
{$ u(t) = \frac{\beta}{\alpha} - Ce^{-\alpha t} $}
However, this is a rather limited case, and this equation does not extend to functional coefficients.
Method of Integrating Factors
This is where the integrating factor comes in. The integrating factor is some function {$\mu(t)$} which, when we multiply both sides of our DE by it, allows us to integrate easily. Our general equation will look like this:
{$ u’(t)\mu(t) + p(t)\mu(t)u(t) = g(t)\mu(t) $}
The next step is realizing the significance of the fact that {$ (u\mu)’ = u’\mu + \mu’u $}. Note that this matches the right hand side of the above equation if {$\mu’(t) = p(t)\mu(t)$}. Note that this is the reason that the exponential function lends itself to solving differential equations with this method - that the derivative of the function must contain the function itself. Using this last fact, when integrating both sides of the general equation, the RHS will simply be {$ u\mu $}, and we simply hope we will be able to integrate the LHS. We can reorder and integrate the above to find that {$\mu(t) = \exp(\int{p(t)dt)}$}. Reordering of our general form with the integrating factor leads us to the following general equations-
{$ u(t) = \frac{\int\mu(t)g(t)dt + C}{\mu(t)} $}, where {$ \mu(t) = \exp(\int p(t)dt) $}
Separable Equations
Sometimes, we get lucky, and can The general form for equations which are best used with this solution method is
{$ M(x) +N(y)\frac{dy}{dx} = 0 $}
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