Separation of Variables

Separation of Variables 1. Separable Equations We will now learn our first technique for solving differential equation. An equation is called separable when you can use algebra to separate the two variables, so that each is completely on one side of the equation. We illustrate with some examples. Example 1. Solve y' = x (y − 1) Solution. We rewrite the equation as variables dy = x (y − 1). Then separate the dx dy = x dx. y−1 Next we integrate both sides. dy = y−1 x dx ⇔ ln |y − 1| + c2 = x2 + c1 . 2 We can amalgamate the two constants of integration into one constant: ln |y − 1| = x2 + c3 . 2 (We label the constant of integration c3 so we’ll have c still available later.) Next we solve for y as a function of x. | y − 1| = e x 2 /2+ c 3 = e c3 e x 2 /2 . The absolute value signs can be removed, but then the right hand side might be positive or negative. We write this as y − 1 = ± e c3 e x 2 /2 ⇔ 1 + ± e c3 e x 2 /2 . Finally we replace the constant ±ec3 by C to get the solution y(t) = 1 + Ce x 2 /2 dy = x dx y−1 might give pause. However, this formal method is justified by the chain rule, in the same way change of variable (u-substitution) is justified for in­ tegration. Note. For the more rigorously minded an expression like Separation of Variables 1.1. OCW 18.03SC Lost Solutions Example 2. Find all the solutions to the DE y' = 2x (1 − y)2 . (1) Solution. First, note there is a constant solution: y( x ) = 1. It is easy to see this is a solution by substituting it into (1) –both sides of the equation become 0. We need to note this because, as we will see, the separation of variables method will not find this particular solution. Now let’s solve the DE by separation of variables. 1. Separate variables: 2. Integrate: 3. Solve for y: dy = 2x dx. (1 − y )2 1 = x2 + C. 1−y 1 . y = 1− 2 x +C Notice that the constant solution y( x ) = 1 is not in the parametrized family found by separation variables. We call this a lost solution because it is lost by separation of variables. How did it get lost? The answer is in step (1) above, where the term dy is only valid if y = 1. (1 − y )2 In general, for the separable DE y' = f ( x ) g(y), all the roots of g(y) give lost (constant) solutions. Example 3. Find all the lost solutions of y' = ( x + 1)e x (y2 − 8y + 7). Solution. The factor y2 − 8y + 7 has roots y = 1 and y = 7. Therefore the lost solutions are the constant functions y( x ) = 1 and y( x ) = 7. 1.2. The Most Important DE Even though we already know the solution, we should test our new technique on the DE for exponential growth/decay. . Example 4. Solve y = ky. Solution. Separate variables: dy = k dt. y Integrate: ln |y| = kt + c1 . Exponentiate: |y| = ekt+C1 = ec1 ekt . 2 Separation of Variables OCW 18.03SC Remove absolute value: y = ±ec1 ekt . Let the constant ±ec1 = C: y = Cekt . All solutions to the DE are y(t) = Cekt . If you look carefully you’ll see we did one rather sneaky thing. The solution y(t) = 0 is a lost solution, yet it appears to have been found by the separation of variables (set C = 0). What happened is that when we renamed ±ec1 as C we should have noticed that the exponential is never 0, so C = 0. Essentially, we included the lost solution by being a little sloppy and then getting lucky. We do not recommend this technique as a way to do mathematics! 3