1. MATH 374 Lecture 7
Linear Equations

2. 2.5: The Linear Equation of Order One
We will now find a way to solve linear equations of order one!
Recall the general form of such an equation:
Dividing (1) by A(x), we can put (1) in standard form:
Now, if we are lucky, putting (2) in the form
will yield an exact equation.

3. What to do if (3) is not exact …
Unfortunately, this usually doesn’t happen.
What we can do is try to find a function v(x) such that multiplying (3) by v(x) yields an exact equation.
With this in mind, multiply (3) by v:
For (4) to be exact, Theorem 2.3 implies we must have
since v, P, and Q are functions of x alone.
Equation (5) is separable and can be solved by integration:

4. Integrating Factor
We only need one function v(x) to make (4) exact, so we choose
(Check that this choice of v(x) works!)
Definition: A function that makes (4) exact is called an integrating factor.

5. How to use an integrating factor …
So now that we can make (4) exact, how does this help us solve (1)?
In practice, the way to solve a first order linear equation is to write it in standard form (2) and multiply by the integrating factor v(x)=es P(x) dx.
This yields

6. General Solution

7. Steps to Solve a Linear First Order Differential Equation
Note: Instead of memorizing (8), it is better to remember the process we just used to get (8):
Put the differential equation in standard form (2):
Find an integrating factor v(x)=es P(x) dx.
Multiply both sides of (2) by the integrating factor.
Realize that this new equation can be written:
Integrate and solve for y.

8. Example 1

9. Existence and Uniqueness
We now look at an existence and uniqueness theorem for linear equations of order one!
Theorem 2.4: For the linear differential equation
dy/dx + P(x) y = Q(x) (10)
if P and Q are continuous functions on a < x < b,
and x = x0 2 (a,b), and if y = y0 is any real number,
then there exists a unique solution y = y(x) of (10) on the interval a < x < b that satisfies the initial condition y(x0) = y0.

10. Proof of Theorem 2.4
Existence: Multiply differential equation (10) by the integrating factor v(x) = es P(x)dx to get
When x = x0, an appropriate choice of C will give y = y0.
Uniqueness: Apply the existence and uniqueness Theorem 1.1 to first order equation (10).

11. Example 2