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Unformatted text preview: Notes on Second Order Linear Differential Equations Stony Brook University Mathematics Department 1. The general second order homogeneous linear differential equation with constant co- efficients looks like Ay 00 + By + Cy = 0, where y is an unknown function of the variable x , and A , B , and C are constants. If A = this becomes a first order linear equation, which we already know how to solve. So we will consider the case A 6 = 0. We can divide through by A and obtain the equivalent equation y 00 + by + cy = where b = B / A and c = C / A . Linear with constant coefficients means that each term in the equation is a constant times y or a derivative of y . Homogeneous excludes equations like y 00 + by + cy = f ( x ) which can be solved, in certain important cases, by an extension of the methods we will study here. 2. In order to solve this equation, we guess that there is a solution of the form y = e x , where is an unknown constant. Why? Because it works! We substitute y = e x in our equation. This gives 2 e x + b e x + ce x = 0. Since e x is never zero, we can divide through and get the equation 2 + b + c = 0. Whenever is a solution of this equation, y = e x will automatically be a solution of our original differential equation, and if is not a solution, then y = e x cannot solve the differential equation. So the substitution y = e x transforms the differential equation into an algebraic equation! 2 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS Example 1. Consider the differential equation y 00- y = 0. Plugging in y = e x give us the associated equation 2- 1 = 0, which factors as ( + 1 )( - 1 ) = 0; this equation has = 1 and =- 1 as solutions. Both y = e x and y = e- x are solutions to the differential equation y 00- y = 0. (You should check this for yourself!) Example 2. For the differential equation y 00 + y- 2 y = 0, we look for the roots of the associated algebraic equation 2 + - 2 = 0....
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