Separable Solutions vs Integrating Factors I am looking to better understand the key differences between these two methods of solving ODEs. I know they are very different, but I will give an example to show what I am a bit confused about.
For example, in one of my previous problems I had $$\frac{dQ}{dt}=3-0.7Q$$
Now, when I first saw this, It appeared to be linear to me. I thought maybe the best way to solve it would be to rewrite it as $$\frac{dQ}{dt}+0.7Q=3$$ and then use the 0.7 to solve for the integrating factor.
However, I was told that the correct answer is to look for separable solutions. Which I also understand how to do. But I am just a bit worried that I will try to solve something by using integrating factors when separation of variables should be used.
Is this the case? How can I know?
Thank you all!
 A: At least three methods can be applied successfully to this equation!
Separating variables is perfectly valid:
$\frac {dQ}{dt}=3-0.7Q$
$\int\frac 1 {3-0.7Q}dQ=\int dt$
$\frac 1 {-0.7} \log\left(3-0.7Q\right)=t+c$
$\log\left(3-0.7Q\right)=-0.7 t+c'$
$3-0.7Q=Ae^{-0.7 t}$
$0.7Q=3-Ae^{-0.7 t}$
$Q=\frac {30} 7-A'e^{-0.7 t}$
Integrating factor is also valid:
$\frac {dQ}{dt}+0.7Q=3$
$\frac {dQ}{dt}e^{0.7t}+0.7Qe^{0.7t}=3e^{0.7t}$
$\frac {d}{dt}\left(Qe^{0.7t}\right)=3e^{0.7t}$
$Qe^{0.7t}=\frac{3}{0.7}e^{0.7t}+B$
$Q=\frac{3}{0.7}+\frac B {Qe^{0.7t}}$
$Q=\frac{30}{7}+ B Qe^{-0.7t}$
Solve first as homogenous DE and then find the particular integral...
Consider first $\frac {dQ}{dt}=-0.7Q$
This has well-known general solution $Q_{general}=De^{-0.7t}$
The non-homogenous part is a constant term, so we have $Q_{trial}(t)=k$ with $\frac d {dt}Q_{trial}(t)=0$.
Need $Q_{trial}$ to satisfy the differential equation.
$0=3-0.7k$
$k=\frac {30} 7$
Combine $Q=Q_{general}+Q_{trial}=De^{-0.7t}+\frac {30} 7$
They all give the same solution - you just have to decide which is more efficient.
With other DEs, maybe only one (or none!) of these methods will be applicable.
A: A separable equation has the form 
$$Q'(t)=f(t)g(Q)$$
while a linear equation has the form
$$Q'(t)+h(t)Q(t)=i(t).$$
When $g(Q)=aQ+b$ is a linear function, you get a separable and linear equation
$$Q'(t)-aQ=f(t)b,$$
and when $i(t)=0$ you get a linear and separable equation
$$Q'(t)=-h(t)Q(t).$$
As you see, there is an overlap between the domain of application of the resolution methods.
Let us compare:


*

*$g(Q)=aQ+b$, separable:


$$\frac{Q'(t)}{aQ+b}=f(t)\to\frac1a{\ln(aQ+b)}=\int f(t)\,dt.$$


*

*$g(Q)=aQ+b$, linear: the integrand factor is $e^{-at}$, and


$$Q'(t)-aQ=f(t)b\to e^{-at}Q(t)=\int e^{-at}f(t)b\,dt.$$
The best option depends on the shape of the function $f$.


*

*$i(t)=0$, separable:


$$Q'(t)=-h(t)Q(t)\to\ln(Q(t))=-\int h(t)\,dt.$$


*

*$i(t)=0$, linear: the integrand factor is $e^{\int h(t)\,dt}$:


$$Q'(t)+h(t)Q(t)=0\to Q(t)e^{\int h(t)\,dt}=C.$$
The two approaches are equivalent.
