$$ \frac{dy}{dx}+Q(x)y=M(x)$$ multiplying with integrating factor $$f(x)\frac{dy}{dx}+f(x)Q(x)y=f(x)M(x) $$ we know that

$$\frac{d(f(x)y)}{dx} = f\cdot \frac{dy}{dx}+\frac{df}{dx}\cdot y$$

  • they both have $y'$ and $y$ but does that immediately make the two equation the same all the time?
  • $\begingroup$ I don't understand the question: the integrating factor, for $\;Q\;$ integrable, is given by $\;f(x)=e^{\int Q(x)dx}\;$ , so the last line follows from this as $\;f'(x) = Q(x)e^{\int Q(x)dx}\;$ ...Does this answer your doubt? $\endgroup$
    – DonAntonio
    Feb 25 at 13:05
  • $\begingroup$ No, what I'm asking is the 2nd equation and the 3rd equation are look alike; therefore, we say that f(x)q(x) = f'. But does they have to be equal, is there no chance of equation being different than d(f(x)y)/dx ? $\endgroup$
    – mark
    Feb 25 at 13:14
  • 1
    $\begingroup$ If you multiplied by an actual integrating factor then no: they must be equal, of course. The third equation is just the left side of the second one... $\endgroup$
    – DonAntonio
    Feb 25 at 13:15

1 Answer 1


Remember the basic product rule:

$$(f g)^\prime(x)=(f g^\prime)(x) + (f^\prime g)(x)$$

The general idea behind introducing an integrating factor in a linear DE is to introduce something that looks like the output of the product rule. So for an expression like

$$I\frac{\mathbf dy}{\mathbf dx}+IQy=IM$$

notice that the left hand side is the output of the product rule, if $\displaystyle\frac{\mathbf d I}{\mathbf dx}=IQ$:

$$\underbrace{I}_{f(x)}\;\underbrace{\frac{\mathbf dy}{\mathbf dx}}_{g^\prime(x)}+\underbrace{IQ}_{f^\prime(x)}\;\underbrace{y}_{g(x)}=IM$$

Thus, anti-differentiating both sides gives us:

$$\underbrace{Iy}_{f(x)g(x)}=\int IM\,\mathbf dx$$

I think that's what you were asking: this is why $\displaystyle\frac{\mathbf d}{\mathbf dx}(Iy)=I\frac{\mathbf dy}{\mathbf dx}+IQy$.

Of course, as already mentioned, this is assuming that $\displaystyle\frac{\mathbf d I}{\mathbf dx}=IQ$. However, the nice thing about $\displaystyle\frac{\mathbf d I}{\mathbf dx}=IQ$ is that it is a separable DE of the form $\displaystyle\frac1I\frac{\mathbf dI}{\mathbf dx}=Q$, which we can solve to get $\displaystyle I=\exp\int Q\,\mathbf dx$. In other words, we introduce a function $I$ with the specific property, that $\displaystyle\frac{\mathbf d I}{\mathbf dx}=IQ$, in order to make the above result true.


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