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Review of Chapter 4 Math 2280 The Chapter 4 exam will focus mostly on higher-order (i.e. second-order and higher) differential equations, generally with constant coefficients. It will help to first review some general theory about these DEs. Let y(k) denote the kth derivative of y. Given an nth order DE an y(n) + an−1 y(n−1) + ... + a1 y0 + a0 y = g(x), (1) the solution y should be written as yc + y p , where yc is the solution if g(x) = 0, and y p is the particular solution when g(x) 6= 0. Of course, if the function g(x) is zero, then y p is zero as well. To find the solution yc to Equation 1, we make the starting assumption that y = emx for some value of m. Taking derivatives gives that y( n) = mn emx . This allows us to find the auxiliary equation an mn + an − 1mn−1 + ... + a1 m + a0 = 0, (2) where the coefficients ai are exactly the same as in Equation 1. So finding the values of m that solve Equation 2 allows us to find the terms in the yc portion to the solution of Equation 1. In particular: • If m is a real root of Equation 2 that is not repeated, then Ci emx is a term in yc . • If m is a real root of Equation 2 that is repeated k times, then Ci x j emx is a term in yc for all values of j from 0 to k − 1. (In other words, for each repetition, multiply by a higher power of x.) • If m = α + iβ is a non-repeated complex root of Equtaion 2, then Ci eαx cos(βx) + Ci+1 eαx sin(βx) are terms in yc . (The number m = α − iβ will also be a root, but these two terms account for both of these values of m.) • If complex roots are repeated k times, then Ci x j eαx cos(βx) +Ci+1 x j eαx sin(βx) appear in yc , where again the values of j start at 0 and run to k − 1. Remember that yc should contain n unknown constants, where n is the same as in Equations 1 and 2. This allows for initial value problems to be solved once the first n − 1 derivatives at a point are given. If one solution y1 is known that solves Equation 1, then another solution can be found by setting y2 = uy1 and substituting into Equation 1. All of the terms that include a u will factor out with an equation that is precisely y1 plugged into Equation 1. As a result, an order (n − 1) differential equation in u can be solved to evaluate uy1 after making the substitution w = u0 . In the case where n = 2, the explicit formula for this is Z − R P(x) dx e dx, y2 = y1 y21 when Equation 1 is written as y00 + P(x)y0 + Q(x)y = 0. (Note that this also works in the case where the coefficients of y are non-constant.) To find y p , there are two main techniques: Undetermined Coefficients and Variation of Parameters. These techniques are only needed when g(x) 6= 0. In both cases, the complementary solution yc should be found first, as parts of these techniques depend on knowing the form of yc . Undetermined Coefficients can only be used when g(x) consists of polynomials, exponential (base e) functions, and the trig functions sine and cosine. The approach is similar to partial fractions. For each term appearing in g(x), take derivatives until no additional functions appear. Then y p is made up of a sum of those functions with coefficients that are solved after plugging y p in to Equation 1 and solving the resulting system. In the event that a term in y p also appears in yc , it should be multiplied by x just as before. Variation of Parameters is a bit more robust, but also a bit more complicated. Here we assume that the particular solution takes the form y p = u1 y1 + u2 y2 + ... + un yn where the yi are all solutions from yc . Then finding all of the individual ui is done by use of the formula u0i = Wi , W where W denotes the determinant of the Wronskian matrix of y1 , y2 , ...yn , and Wi is the determinant of the matrix W with column i replaced by a column of zeros, with g(x) as the last entry. (Remember that the idea here comes from Cramer’s Rule.) In particular, with n = 2 we have 0 g(x) u01 = y1 y01 y2 y02 −y2 g(x) , = W (y1 , y2 ) y2 y02 y1 0 y01 g(x) y1 g(x) 0 u1 = . = W (y1 , y2 ) y1 y2 y01 y02 With n = 3, 0 0 g(x) u01 = y1 y01 y001 y2 y02 y002 y2 y02 y002 y3 y03 y003 , y3 y03 y003 y1 y01 y00 u02 = 1 y1 y01 y001 0 0 g(x) y2 y02 y002 y3 y03 y003 , y3 y03 y003 y1 y01 y00 u03 = 1 y1 y01 y001 y2 y02 y002 y2 y02 y002 0 0 g(x) . y3 y03 y003 Note that these are all solved for u0i , so an integral still needs to be taken to find the actual ui . The +C at the end of each of these can be taken to be anything, so setting C = 0 is often the choice made. Lastly, these methods can all be applied to systems of differential equations, wherein we assume we have functions x = x(t), y = y(t), and sometimes even z = z(t). In these cases, it helps to think of derivatives as being a “coefficient” D that can be factored out. (So (D + 3)y = Dy + 3y = y0 + 3y.) This allows the elimination technique from Algebra to be used to reduce to a differential equation of just one function that can be solved using the above techniques. Then back-substitution can be used to find the other functions. In each of these cases, both the complementary and particular solutions should be considered. Other Things: • Just like on the Chapter 2 exam, integration techniques are vital. Make sure you’re comfortable with the ones that have been on the homework. • Some problems have multiple techniques that can work. Any valid technique will receive credit. Suggested review problems: 1. Solve: y00 + 4y0 − y = 0. 2. Solve: y000 − 4y00 − 5y0 = 0. 3. Solve: y(4) − 7y00 − 18y = 0. 4. Solve: y00 − 2y0 + y = 0, y(0) = 5, y0 (0) = 10. 5. Solve: y000 + 12y00 + 36y0 = 0, y(0) = 0, y0 (0) = 1, y00 (0) = −7. 6. Given that y1 = ln(x) is a solution to xy00 + y0 = 0, find a second solution y2 . 7. Given that y1 = 1 is a solution to (1 − x2 )y00 + 2xy0 = 0, find a second solution y2 . 8. Given that y1 = x sin(ln x) is a solution to x2 y00 − xy0 + 2y = 0, find a second solution y2 . 9. Solve: y00 − y0 = −3 10. Solve: y00 − 16y = 2e4x 11. Solve: y00 − 2y0 + 5y = ex cos(2x) 12. Solve: y(4) − y00 = 4x + 2xe−x 13. Solve: y00 + 4y = cos(2x), y(0) = y0 (0) = 0 14. Solve: y00 + y = cos2 (x) 15. Solve: y00 − 9y = 9x e3x 16. Solve: y00 − 2y0 + y = 17. Solve: 18. Solve: 19. Solve: ex 1+x2 dx dt dy dt = −y + t = x − t (D + 1)x + (D − 1)y = 2 3x + (D + 2)y = −1 (D − 1)x + (D2 + 1)y = 1 (D2 − 1)x + (D + 1)y = 2