4.4 Solving Initial Value Problems

Transcription

1 4.4. SOLVING INITIAL VALUE PROBLEMS 4.4 Solving Initial Value Problems 4.4. Description of the Method and Examples In the introduction of the previous section, we used an example to show how the Laplace transform could be used to solve initial value problems. We summarize the steps involved in this technique, then look at some examples. In this outline, we assume that in the equation we are solving, the independent variable is t, the dependent variable is y = y (t) and we denote the Laplace transform of y by Y (s) = L {y (t)} (s). In other words, y (t) = L {Y (s)} (t). Generally speaking, this technique can be seen as a change of variable. Given an equation where the unknown function is y (t), using the Laplace transform, we rewrite it as an equation in Y (s), the Laplace transform of y (t). In other words, Y (s) = L {y (t)} (s). If we can solve this new equation for Y (s), then we can find y (t) since y (t) = L {Y (s)} (t). More specifically, To solve an initial value problem involving an unknown function y (t), using the Laplace transform:. Take the Laplace transform on both sides of the equation. 2. Using the properties of the Laplace transform and the initial conditions, obtain an equation in Y (s). 3. Solve this equation for Y (s). 4. Find y (t) = L {Y (s)} (t) using the techniques described in the previous section. Remark 4.4. One of the formulas that will be used by this technique is the formula of the Laplace transform of the derivatives of a function, L { f (n)} (s) = s n L {f} (s) s n f (0) s n 2 f (0) s n 3 f (0)... f (n ) (0). Note that this formula uses the value of the function in question and its derivatives at t = 0. This implies that the initial conditions of our initial value problem must be at t = 0. We will show how to handle the case when the initial condition is not at t = 0 in one of the examples. Example Solve the initial value problem. Take the Laplace transform. L {y 2y + 5y} (s) = L { e t} (s) y 2y + 5y = e t y (0) = 2 y (0) = 2

2 2 CHAPTER 4. LAPLACE TRANSFORMS 2. Obtain an equation in Y and solve the above equation for Y L {y } (s) 2L {y } (s) + 5Y = L { e t} (s) s 2 Y (s) sy (0) y (0) 2 (sy (s) y (0)) + 5Y (s) = s + s 2 Y (s) 2s 2 2sY (s) Y (s) = s + Y (s) ( s 2 2s + 5 ) 2s 8 = s + Y (s) ( s 2 2s + 5 ) = s + + 2s + 8 Y (s) ( s 2 2s + 5 ) = 2s2 + 0s (s + ) Y (s) = 2s 2 + 0s (s + ) (s 2 2s + 5) 3. Find y (t) = L {Y (s)} (t) In an example in the previous section, we found that { L 2s 2 } + 0s (s 2 = 3e t cos 2t + 4e t sin 2t e t 2s + 5) (s + ) hence y (t) = 3e t cos 2t + 4e t sin 2t e t Remark Of course, this is an equation with constant coeffi cient, which we could have solved using other techniques we have studied. We solved it using the Laplace transform to illustrate this technique. Example Solve the initial value problem y + 4y 5y = te t y (0) = y (0) = 0. Take the Laplace transform. L {y + 4y 5y} = L { te t}

3 4.4. SOLVING INITIAL VALUE PROBLEMS 3 2. Obtain an equation in Y and solve the above equation for Y L {y } + 4L {y } 5Y = s 2 Y sy (0) y (0) + 4 (sy y (0)) 5Y = 3. Find y (t) = L {Y (s)} (t) s 2 Y s + 4sY 4 5Y = Y ( s 2 + 4s 5 ) = (s ) 2 (s ) 2 (s ) 2 (s ) s Y (s ) (s + 5) = s3 + 2s 2 7s + 5 (s ) 2 Y = s3 + 2s 2 7s + 5 (s ) 3 (s + 5) Before we find the inverse Laplace transform, we need to decompose s3 + 2s 2 7s + 5 (s ) 3 (s + 5) into partial fractions. s 3 + 2s 2 7s + 5 (s ) 3 (s + 5) = = A s + B (s ) 2 + C (s ) 3 + D s (s ) 36 (s ) (s ) (s + 5) Hence, { } L 8 26 (s ) 36 (s ) (s ) (s + 5) = 8 { } { } { } 26 L 36 s L (s ) L (s ) { } 26 L s + 5 We compute each of these inverse Laplace transforms and put everything together. { } L = e t s { } L (s ) 2 = te t { } L (s ) 3 = 2 t2 e t

4 4 CHAPTER 4. LAPLACE TRANSFORMS { } L = e 5t s + 5 So y (t) = 8 26 et 36 tet + 2 t2 e t e 5t w (t) 2w (t) + 5w (t) = e π t Example Solve the initial value problem w (π) = 2 w (π) = 2 We need the initial condition to be at t = 0 because of the formula for the Laplace transform of derivatives. We can fix this with the change of variable y (t) = w (t + π). With this, w (π) = y (0). Also, y (t) = w (t + π) and y (t) = w (t + π). So w (π) = y (0). Since the original equation is valid for any t, replacing t by t + π gives w (t + π) 2w (t + π) + 5w (t + π) = e t which becomes y (t) 2y (t) + 5y (t) = e t and the initial conditions are y (0) = 2 and y (0) = 2. At this point, the Laplace transform technique can be used. We already solved this problem in the first example. We found that y (t) = 3e t cos 2t + 4e t sin 2t e t. We need to give the solution in terms of w. Since y (t) = y (t) = w (t + π) it follows that w (t + π) = 3e t cos 2t + 4e t sin 2t e t hence w (t) = 3e t π cos 2 (t π) + 4e t π sin 2 (t π) e t+π we replaced t by t π. We can simplify this further, using the fact that sine and cosine are periodic of period 2π hence sin (t ± 2π) = sin t and cos (t ± 2π) = cos t. It follows that w (t) = 3e t π cos 2t + 4e t π sin 2t e t+π y + 2ty 4y = Example Solve the initial value problem y (0) = 0 y (0) = 0. Take the Laplace transform. L {y + 2ty 4y} = L {} 2. Obtain an equation in Y and solve the above equation for Y L {y } + 2L {ty } 4Y = ( s s 2 Y sy (0) y (0) + 2 d ) ds (L {y }) 4Y = s s 2 Y sy (0) y (0) + 2 ( dds ) (sy y (0)) 4Y = s s 2 Y 2 d ds sy (s) 4Y = s s 2 Y 2sY 2Y 4Y = s 2sY + ( s 2 6 ) Y = s

5 4.4. SOLVING INITIAL VALUE PROBLEMS 5 This is a linear equation, its general form is ( 3 Y + s s ) Y = 2 2s 2 The integrating factor is µ (s) = e ( 3 s s 2)ds = e 3 ln s s2 4 = s 3 e s2 4 Multiplying the equation above by the integrating factor gives us Hence d ) (s 3 e s2 4 Y = s ds s 3 e s2 4 Y = s With u = s2 4, du = s 2 ds so s follows that 2 e s 2 s 3 e s2 4 Y = e s2 4 2 e s e s 2 4 ds 4 ds = e u du = e u = e s2 4. It + C or Y = s 3 + Ce s 2 4 s 3 In one of the homework problems, we proved that if F (s) = L {f (t)} (s) and f is of exponential order, then lim F (s) = 0 hence, we must have s Ce s2 4 lim Y (s) = lim s s s 3 = 0. It follows that C = 0 and Y (s) = s Find y (t) = L {Y (s)} (t) y (t) = L {Y (s)} (t) = L { s 3 } (t) = 2 L { 2 s 3 } (t) = 2 t Exercises Do numbers, 3, 5, 7, 9,, 5, 7, 9 at the end of section 7.5 in your book.

6 Bibliography [] Paul Blanchard, Robert L. Devaney, and Glen R. Hall, Diff erential equations, fourth ed., Brooks/Cole, CENGACE Learning, 202 (English). [2] Charles H. Edwards, David E. Penney, and David T. Calvis, Diff erential equations and boundary value problems: Computing and modeling, fifth ed., Pearson, 205 (English). [3] R. K. Nagle, Edward B. Saff, and Arthur D. Snider, Fundamentals of differential equations, eigth ed., Pearson/Addison-Wesley, 202 (English). [4] Virginia W. Noonburg, Ordinary diff erential equations: from calculus to dynamical systems, The Mathematical Association of America,