Integrating Factor

Use the Product Rule to convert the DE into a form where we can apply Direct Integration

t^{2} y^{'} + 2 t y = e^{2 t} \to (t^{2} y)^{'} = e^{2 t}

For DEs where the Product Rule cannot be applied directly, we can multiply every term by a function of t where product rule works

need $\frac{d}{d t} μ (t) = 2 μ (t)$ in this case

y^{'} + 2 y = 5 \to μ (t) y^{'} + 2 μ (t) y = 5 μ (t)

$μ$ in this case will be $μ = c e^{2 t}$ after integrating, so any of those possible integrating factors will get the DE in the form required to use Product Rule

e^{2 t} y^{'} + 2 e^{2 t} y = 5 e^{2 t} \to (e^{2 t} + y)^{'} = 5 e^{2 t}

And then we can Integrate with respect to t:

e^{2 t} y = \int 5 e^{2 t} d t = \frac{5}{2} e^{2 t} + C

So $y = \frac{5}{2} e^{2 t} \frac{1}{e^{2 t}} + \frac{C}{e^{2 t}} = \frac{5}{2} + C$

In general, for a first order linear DE in standard form

y^{'} + p (t) y = g (t)

Will always allow us to find an integrating factor $μ (t)$ as

μ (t) = e^{\int p (t) d t}

Exact Equations