# Section 3.6: Nonhomogeneous 2 nd Order D.E.s Method of Undetermined Coefficients Christopher Bullard MTH-314-001 5/12/2006.

## Presentation on theme: "Section 3.6: Nonhomogeneous 2 nd Order D.E.s Method of Undetermined Coefficients Christopher Bullard MTH-314-001 5/12/2006."— Presentation transcript:

Section 3.6: Nonhomogeneous 2 nd Order D.E.s Method of Undetermined Coefficients Christopher Bullard MTH-314-001 5/12/2006

Nonhomogeneous Equations: Assumptions Form: L(y)= y + p(t)y + q(t)y = g(t), where g(t) is not equal to zero. p(t) and q(t) are continuous for all t in the domain. Now consider the homogeneous equation y + p(t)y + q(t)y =0 If we consider the only difference between the equations as their end result, g(t), we can claim…

Theorem 3.6.1 If Y 1 and Y 2 are solutions to L(y), then Y 1 - Y 2 is a solution to the homogeneous equivalent of L(y) (where g(t) = 0). This means if y1 and y2 are solutions to the homogeneous equation, then Y 1 - Y 2 = c 1 y 1 (t) + c 2 y 2 (t) How does this help us solve the nonhomogeneous equation? Consider the following proof…

Proof: Assume Y 1 and Y 2 meet the following requirements: L[Y 1 ](t) = g(t) and L[Y 2 ](t) = g(t) Subtracting the two, we find… L[Y 1 ](t) - L[Y 2 ](t) = g(t) – g(t) = 0 Using the distributive property… L[Y 1 (t) - Y 2 (t)]= L[Y 1 ](t) - L[Y 2 ](t) Therefore, L[Y 1 (t) - Y 2 (t)]= c 1 y 1 (t) + c 2 y 2 (t), which leads us to our next theorem…

Theorem 3.6.2: A solution to the nonhomogeneous equation L(y) is of the form… y(t)= c 1 y 1 (t) + c 2 y 2 (t) + Y(t) Remember c 1 y 1 (t) + c 2 y 2 (t) are the constants and solutions to the homogeneous equation. Y(t) is the nonhomogeneous solution, while y(t) is the complete solution to the general nonhomogeneous equation.

Steps to find the solution Y(t): Find c 1, y 1, c 2, and y 2. Find a particular solution Y(t) to the nonhomogeneous equation (anything that solves y + p(t)y + q(t)y = g(t)). Add Y(t) to c 1 y 1 (t) + c 2 y 2 (t), and then solve for the initial conditions. But where do undetermined coefficients come into play?

Undetermined Coefficients: Uses Advantages: Very easy procedure (assuming a good guess) Disadvantages: Difficult to guess Y(t) Limitations on the number of equations we can solve based on the form of g(t). So, which forms of g(t) can we solve for with the Undetermined Coefficients Method? Polynomials Exponential (e rt ) functions Trigonometric functions with sin(x) and cos(x)

Potential Problems: If Y(t) results in the same solution to the homogeneous equation (where g(t) = 0), we need to find another solution. Usually, the answer is to multiply Y(t) by t. If this does not work, multiply by t again (to get t 2 attached to Y(t)). Why does this work? From Section 2.1, we could solve first order differential equations with a similar problem (y – y = 2e -t ). We find y(t) in this case was 2te -t. We just use this similar case as a template for our approach to cases when Y(t) = y 1 or y 2.

Summary of Procedures: Find the homogeneous solution c 1 y 1 (t) + c 2 y 2 (t). If g(t) is a polynomial, exponential, or trigonometric function with sin(x) or cos(x), we can find Y(t). What about combinations of the above? Simply find Y(t) for the individual cases, and algebraically add them together. Add the homogeneous and nonhomogeneous solutions. Use the initial conditions to find the particular solution.

Other Examples and Sources of Information: S.O.S. Math: Method of Undetermined Coefficients Paul's Online Math Notes (Lamar University) Differential Equations Notes ( Bruce Ikenaga of Millersville University)Differential Equations Notes ( Bruce Ikenaga of Millersville University) Work Cited: Boyce, William E., DiPrima, Richard C. Elementary Differential Equations: Eighth Edition. New York:John Wiley and Sons, Inc. 2005

Similar presentations