MATH 374 Lecture 7 Linear Equations

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.

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:

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.

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

General Solution

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.

Example 1

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.

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).

Example 2