Superposition Principle

Superposition Principle 1. Superposition Principle for Inputs We conclude our introduction to first order linear equations by dis­ cussing the superposition principle. This is the most important property of these equations. In fact, we will see in later sessions that superposition is the defining characteristic of linear equations of any order. In the examples below we will see that the superposition principle al­ lows us to break up a problem into simpler problems and then at the end assemble the answer from its simpler pieces. As usual in this session we are considering the first order linear equa­ tion . y + p ( t ) y = q ( t ). We will consider the left-hand side to be the system and the right-hand side to be the input. For any given input q(t) that has output y(t) we will write q � y (read input q leads to output y). If q1 (t) and q2 (t) are signals, and c1 and c2 are constants then we call c1 q1 (t) + c2 q2 (t) a superposition of q1 and q2 . Another name we will often use for this is a linear combination of q1 and q2 . Superposition Principle: For constants c1 and c2 If q1 � y1 and q2 � y2 then c1 q1 + c2 q2 � c1 y1 + c2 y2 . Proof: This is true because the ODE is linear. The proof takes two lines: . . d ( c1 y1 + c2 y2 ) + p(c1 y1 + c2 y2 ) = c1 (y1 + py1 ) + c2 (y2 + py2 ) dt = c1 q1 + c2 q2 . We present some easy examples below. 2. Examples In the next session we will learn how to find the solutions to the follow­ ing ODE’s . For now, you can check the given solutions by substitution into the DE. . i. x + 2x = 1 has a solution x (t) = 1 2 Superposition Principle OCW 18.03SC . ii. x + 2x = e−2t has a solution x (t) = te−2t . iii. x + 2x = 0 has a solution x (t) = e−2t . Using the solutions above as a basis, we can solve more complicated equations. . Example 1. Use superposition to find a solution to x + 2x = 1 + e−2t Solution. The input is a superposition of the inputs from (i) and (ii). There­ fore a solution is x (t) = 12 + te−2t . . Example 2. Find a solution to x + 2x = 2 + 3e−2t . Solution. The input is 2 · (1) + 3 · (e−2t ); it is a superposition (with coef­ ficients c1 = 2, c2 = 3) of the inputs from (i) and (ii). Therefore, x (t) = 2 · 12 + 3(te−2t ) = 1 + 3te−2t is a solution. . Example 3. Find lots of solutions to x + 2x = 1 Solution. We can write the input as: 1 = 1 + c · 0. That is, as a superposition of the input from (i) and the homogeneous equa­ tion (iii). Therefore x (t) = 12 + ce−2t is a solution for any value of the pa­ rameter c. . Example 4. Find lots of solutions to x + 2x = 1 + e−2t . Solution. Use superposition: x (t) = 1 2 2 + te−2t + ce−2t .