Solutions to Linear First Order ODE’s

Solutions to Linear First Order ODE’s 1. First Order Linear Equations In the previous session we learned that a first order linear inhomogeneous ODE for the unknown function x = x (t), has the standard form . x + p ( t ) x = q ( t ). (1) (To be precise we should require q(t) is not identically 0.) We saw a bank example where q(t), the rate money was deposited in the account, was called the input signal. We also saw an RC circuit example where the input signal was the voltage V (t) and q(t) = V � (t). A first order linear homogeneous ODE for x = x (t) has the standard form . x + p(t) x = 0. (2) We will call this the associated homogeneous equation to the inhomoge­ neous equation (1) In (2) the input signal is identically 0. We will call this the null signal. It corresponds to letting the system evolve in isolation without any external ’disturbance’. • In the bank example: if there are no deposits and no withdrawals the input is 0. • In the RC circuit example: if the power source is turned off and not providing any voltage increase then the input is 0. 2. Solutions to the Homogeneous Equations The homogeneous linear equation (2) is separable. We can find the so­ lution as follows: dx = − p(t)dt. x • Separate variables: • Integrate: ln | x | = − • Exponentiate: � p(t)dt + c1 . (We use c1 to save C for later.) | x | = e c1 e − � p(t)dt . Solutions to Linear First Order ODE’s • Rename ec1 as C: | x | = C e− OCW 18.03SC � p(t)dt ; C > 0. • Drop the absolute value and recover the lost solution x (t) = 0: This gives the general solution to (2) x (t) = C e− � p(t)dt where C = any value. (3) A useful notation is to choose one specific solution to equation (2) and call it xh (t). Then the solution (3) shows the general solution to the equation is x (t) = Cxh (t). (4) There is a subtle point here: formula (4) requires us to choose one solution to name xh , but it doesn’t matter which one we choose. We can say this somewhat awkwardly as choose an arbitrary specific solution.’ A typical choice is to set the parameter C = 1, but this is not necessary. . Example. Solve x + 2tx = 0. Solution. • Separate variables: • Integrate: dx = −2tdt. x ln | x | = − � 2tdt = −t2 + c1 . • Exponentiate and substitute C for ec1 : 2 2 | x | = e c1 e − t = C e − t . • Drop the absolute value and also recover the lost solution: 2 In this example an obvious choice for xh is xh (t) = e−t . It is clear the general solution to the example is x (t) = C xh (t) where C = any number. 3. Solution to Inhomogeneous DE’s Using Integrating Factors We start with the integrating factors formula: the general solution to the . inhomogeneous first order linear ODE (1) (x + p(t) x = q(t)) is �� � � 1 x (t) = u(t)q(t)dt + C , where u(t) = e p(t) dt . (5) u(t) 2 2 x ( t ) = C e−t . Solutions to Linear First Order ODE’s OCW 18.03SC The function u is called an integrating factor. This method, due to Euler, is easy to apply. We deduce it by the method of optimism, i.e., we introduce an integrating factor u and hope that it will help us. Proof: We start with the product rule for differentiation . . d (ux ) = ux + ux. dt and the equation (1): . x + p ( t ) x = q ( t ). Multiply both sides of the equation by some function u(t), whose value we will determine later: . ux + upx = uq. (6) In order to be able to apply the product rule we want the sum on the left . . hand side of the equation to have the form dtd (ux ) = ux + ux. There may be many functions u for which the left hand side has this form; we only need to find one of them. To do this, note that . . . . . d (ux ) = ux + upx ⇔ ux + ux = ux + upx ⇔ u = up. dt The last equation is a separable DE for the unknown function u: du = p(t) dt u and so: ln |u| = � u=e � p(t) dt p dt . (7) Remember, we are looking for just one u, so any choice of anti-derivative of p(t) in equation (7) will do. Now replace the left-hand side of (6) by d dt ( ux ) and solve for x: . ux + upx = uq d (ux ) = uq dt � u(t) x (t) = u(t)q(t)dt + c �� � 1 x (t) = u(t)q(t)dt + c u(t) 3 Solutions to Linear First Order ODE’s OCW 18.03SC This last equation is exactly the formula (5) we want to prove. Example. factors. . Solve the ODE x + 2x = e3t using the method of integrating Solution. Until you are sure you can rederive (5) in every case it is worth­ while practicing the method of integrating factors on the given differential equation. (At the end, we will model a solution that just plugs into (5).) Multiply both sides by u: . ux + 2u(t) x (t) = u(t) · e3t . Next, find an integrating factor u so that the left-hand side is equal to . . (which equals ux + ux). . . (8) d dt ( ux ) . ux + ux = ux + 2ux ⇒ . u = 2u u(t) = e2t (we choose any one u that works). Now substitute u(t) = e2t into (8), then replace the left-hand side by and solve for x. d 2t (e x ) = e2t e3t dt 1 ⇒ e2t x = e5t + C (integrate the previous equation) 5 1 ⇒ x (t) = e3t + Ce−2t (solve for x (t)). 5 Here is a model of the same solution using (5) directly. Integrating factor: u(t) = e Solution: � 2 dt = e2t (choose any one possibility). 1 x (t) = u(t) =e −2t =e −2t � u(t)e3t dt � e5t dt � 1 5t e +C 5 1 = e3t + Ce−2t . 5 4 � d dt ( ux ) Solutions to Linear First Order ODE’s OCW 18.03SC 4. Comparing the Integrating Factor u and xh Recall that in section 2 we fixed one solution to the homogeneous equa­ tion (2) and called it xh . The formula for xh is xh (t) = e− � p(t) dt , where we can pick any one choice for the antiderivative. Comparing this with the formula for the integrating factor u=e � p(t)dt we get the following relationship between the two functions: xh (t) = 1 . u(t) The solution to the homogeneous equation (or for short the homogeneous solution) xh will play an extremely prominent role in the rest of the course. 5