A 1st order de is linear if it can be expressed as
\(y'+p(x)y=f(x)\) \(\leftarrow \) “standard form"
for some functions \(p(x),f(x)\); otherwise it’s called non-linear.

A 1st order linear de that can be written in the form \(y'+p(x)y=0\) is called homogeneous; otherwise it is called non-homogeneous.

Note: \(y=0\) is a solution to any linear homogeneous 1st order de, called the trivial solution.

For example, \(xy'=x^2-y\cos {(x)}\) is linear as it can be written as \(\displaystyle y'+\frac {\cos {(x)}}{x}y=x\). But the equation \(xy'=x^2-x\cos {(y)}\) is not linear as we cannot make look it \(y'+p(x)y=f(x)\).

Consider the linear homogeneous equation \(y'+p(x)y=0\).

Either \(y=0\) or \(y\neq 0\).

Let’s consider the case where a solution \(y\) is non-trivial. Since \(y\) is differentiable, it must be continuous and thus there is an open interval \(I\) on which \(y\) never has \(0\) as an output.

Then on \(I\),

\(y'=-p(x)y\) by subtracting \(p(x)y\) from both sides. Dividing both sides by \(y\),

\(\displaystyle \frac {1}{y} y'=-p(x)\) on \(I\). By integrating both sides

\(\ln {|y|}=-P(x)+C\) where \(P(x)\) is any antiderivative of \(p(x)\) and \(C\) is an arbitrary constant.

So \(|y|=e^{-P(x)+C}=e^C e^{-P(x)}\).

Then \(y=\pm e^C e^{-P(x)}\).

Relabeling \(\pm e^C\) (arbitrary non-zero constant) by \(C\), and even allowing \(C=0\), gives all solutions:

\(y=C e^{-P(x)}\) (Such a family of functions is the \(``\)general solution" to a de.)

Note: \(I\) could be any open interval where \(p(x)\) is continuous, even \((-\infty ,\infty )\) if \(p(x)\) is continuous everywhere, because if \(C\neq 0\) then \(y\) is never equal to \(0\).

Now consider the linear non-homogeneous equation \(y'+p(x)y=f(x)\) where \(f(x)\neq 0\).

We would like to find the general solution to this de. Observe that \(y=0\) is NOT a solution.

Method of integrating factors: Find an integrating factor \(\mu (x)\), which is a function satisfying...

\(\mu (x)\, y'+ \mu (x) p(x)\, y=[\mu (x)\, y]'\).

Equivalently,

\(\mu (x)\, y'+ \mu (x) p(x)\, y=\mu (x) \,y'+\mu '(x) \,y\).

Then \(\mu \) must satisfy the linear homogeneous equation \(\mu '=p(x)\, \mu \).

Since we only need a solution we pick \(\mu =e^{P(x)}\) where \(P(x)\) is any antiderivative of \(p(x)\). Memorize this!

So why do we call it an integrating factor (I.F.)? Well:

(A) Multiply both sides of \(y'+p(x)y=f(x)\) by \(\mu \) to get \(\mu (x)\, y'+ \mu (x) p(x)\, y=\mu (x)f(x)\).

(B) Use that \(\mu \) is an integrating factor to rewrite this as \([\mu (x)\, y]'=\mu (x)f(x)\).

(C) Integrate both sides to get \(\displaystyle \mu (x)\, y=\int \mu (x)f(x)\;dx\). Divide by \(\mu \).

Then \(\displaystyle y =\frac {1}{\mu (x)} \int \mu (x)f(x)\;dx\). \(\leftarrow \) You could choose to memorize, but I prefer going through the steps!

Notes:

-The possible open intervals for the solutions depend on the functions \(p(x)\) and \(f(x)\).

-Often we drop the \(``\)of x" notation and write \(\displaystyle y=\frac {1}{\mu } \int \mu f \;dx\).

Example: Find the general solution to \(y'=\frac {2y}{x}+x\) on \((0,\infty )\).

First write it in standard form: \(y'-\frac {2}{x}y=x\).

So an integrating factor is \(\mu =e^{-2\ln {|x|}}= e^{\ln {x^{-2}}}=x^{-2}=\frac {1}{x^2}\).

Mult. both sides of the de by \(\mu \):

\(\frac {1}{x^2}\,y' -\frac {2}{x^3}\, y=\frac {1}{x}\) , so:

\(\left (\frac {1}{x^2}\,y \right )'=\frac {1}{x}\). By integrating \(\frac {1}{x^2}\,y=\ln {(x)}+C\).

Hence \(y=x^2\ln {(x)}+Cx^2\) is the general solution on \((0,\infty )\).

A direction field with a solution curve. [Picture]

Figure 1: A direction field with a solution curve