# Need to understand how the integrating factor is found

Given

\begin{align} y &= \frac{a}{b} \left(1 - e^{-bx} \right) \\ \implies y’ &= \frac{a}{b} \cdot b e^{- bx} \\ &= ae^{- bx} \\ \implies y’ &= a \left(1 - \frac{by}{a} \right) \\ &= a - by \end{align}

Perfectly logical. No problem at all.

But I have no clue at all how to go in reverse from

$$y’ = a - by$$

to the solution $$y$$. I think I need to find an integrating factor, but how? I also thought about a characteristic equation. Is that the way to go?

This is probably an obvious question, but all I know about differential equations, I picked up on my own rather than from a course or a text. I understand separable equations, exact equations, and characteristic equations, but finding integrating factors is terra incognita.

• $y'=a-by$ is separable $$\frac{dy}{a-by}= dx \implies \ln (a - by) = - b x + C \implies \dots$$ or can be integrated using an integrating factor as you stated $$y' + by = a \implies \text{IF} = \exp \left(\int b dx \right) = \exp \left( bx\right) \implies e^{bx} y' + be^{bx}y = ae^{bx} \implies \dots$$ or can be solved by finding the homogeneous (by making the ansatz $y_{h} = e^{\lambda x}$ and deriving the characteristic equation) and particular ($y_{b}=A=\text{constant}$) solutions. Jun 26 at 20:31
• Thank you very much. I had a blind spot on the separabilty; I would have wasted hours on that. I shall have to think about the integrating factor logic. But again, thank you. Jun 26 at 20:43
• Oh, and thanks as well to the edits to put the question in LaTeX. I tried quite a few delimiters to invoke it, but none worked. Jun 26 at 21:08
• The integrating factor is simply $e^{bx}$ Jun 26 at 21:30
• Ahh, the coefficient of y becomes the coefficient of x in the exponential. Always obvious once the light bulb turns on. Thank you Jun 26 at 22:57

If you have an ODE of the form $$y' + fy = g,$$ where $$f$$ and $$g$$ are given functions, then the integrating factor is $$e^{F},$$ where $$F$$ is a primitive function of $$f,$$ i.e. $$F'=f$$: $$e^F(y'+fy) = (e^Fy)'^.$$