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Math 217: Differential Equations Final Review Mark Pedigo 1 First-Order DEs Background Info Definition 1.1 (Differential Equation). A differential equation (abbreviated DE) is an equation relating an unknown function and its derivatives. Note: The solution is not a number, as in algebra, but a function. Methods of Solution Separable Equations • Solution Strategy: Put all y terms on one side of the equation, all x terms on the other and integrate. Example 1.2. Solve the DE: dy dx = y 2 −1 ; x y(1) = 2. 2 = y x−1 implies Z Z 1 1 dx = dy x y2 − 1 Z Z 1 1 1 1 =− dy + dy (by partial fractions). 2 y+1 2 y−1 Cx2 +1 1 Thus, 2 ln x + C = ln y−1 , so Cx2 = y−1 . Consequently, y = 1−Cx 2. y+1 y+1 Solution. dy dx Now, use the initial conditions to find C. We’re given y(1) = 2, so 2 = consequently C = 13 . Therefore, y= 1 1/3 x2 + 1 3 + x2 = . 1 − 1/3 x2 3 − x2 Don’t worry too much about absolute value – the sign washes out in C. 1 (1) (2) C+1 1−C and Linear First-Order Equations y 0 + P (x)y = Q(x) • Solution Strategy: Reduce to separable equation by multiplying through by an integrating factor. • Procedure 1. Calculate the integrating factor R ρ(x) = e P (x) dx 2. Multiply both sides of the DE by ρ(x) dy + P (x)y = ρ(x)Q(x) ρ(x) dx d 3. The left-hand side should equal dx [ρ(x)y]. (Mentally check this to make sure you haven’t erred.) Then equation is now d [ρ(x)y] = ρ(x)Q(x) dx 4. Integrate both sides with respect to x. Z ρ(x)y = ρ(x)Q(x) dx. Now, solve for y. Example 1.3. Solve the DE: y 0 + 2xy = x, y(0) = 2. Solution. The steps to find a solution are as follows. 1. Find the integrating factor: ρ(x) = e R 2x dx = ex 2 2 2 2. Multiply both sides by the integrating factor: ex (y 0 + 2xy) = ex · x h i h i 2 2 2 2 d d 3. By the product rule, dx ex y = ex (y 0 + 2xy). Therefore, dx ex y = xex 4. Integrate both sides: x2 e y= 2 Therefore, y = e−x 1 x2 e 2 Z 2 xex dx (Substitute u = x2 ) 1 2 = ex + C. 2 2 + C = 12 + Ce−x . 2 Substitutions • Solution Strategy: Use an appropriate substitution to reduce the DE to one we know how to solve. 1. Homogeneous First-Order Linear DE dy = F xy dx dy vx, dx =v+ • A first-order linear DE that can take the form y x • Solution Strategy: Substitute v = (so y = . dv x dx ). dy Example 1.4. Solve: 2xy dx = 4x2 + 3y 2 . Solution. Dividing through by 2xy yields dy x 3 y =2· + · . dx y 2 x Let v = xy . Then y = vx and yields dy dx = v+x v+x Solving for x dv dx dv . dx (3) Plugging the substitutions into (3) 1 3 dv =2· + v dx v 2 gives us x dv 2 3 = + v−v dx v 2 2 1 = + v v 2 4 + v2 = . 2v This equation is separable. (That was our goal.) We separate variables and integrate both sides: Z Z 1 2v dx = dv. 2 x v +4 This implies ln x + C = ln(v 2 + 4), so Cx = v 2 + 4. Finally, we substitute in for v: y2 Cx = 2 + 4. x Thus, a solution (given in implicit form) is y 2 + 4x2 = Cx3 . 2. Bernoulli Equation 3 y x back • dy dx + P (x)y = Q(x)y n . • Solution Strategy: Use a substitution to reduce the Bernoulli DE to a first-order linear equation in v. Solve this, back-substitute y. • Procedure: Substitute v = y 1−n (so y = v 1/(1−n) , Example 1.5. Solve: dy dx − 3 y 2x = dy dx = dy dv · dv ). dx 2x . y dy 3 − 2x y = 2xy −1 . Then n = −1, so we let Solution. Rewrite the given DE as dx dy dy dv dv v = y 1−(−1) = y 2 . Then y = v 1/2 and dx = dv · dx = 12 v −1/2 dx . Substituting these expressions into our DE yields 1 −1/2 dv 3 v − v 1/2 = 2xv −1/2 . 2 dx 2x Multiplying through by 2v 1/2 yields 3 dv − v = 4x, dx x a first-order linear DE in v. Solving this equation using an appropriate integrating factor yields v = −4x2 + Cx3 . Finally, substituting v = y 2 back into the equation gives us the solution (in implicit form) y 2 = −4x2 + Cx3 . 3. Reducible Second-Order DEs • Reduce to first-order linear DE There are two special cases of second-order DEs which can be reduced to first-order DEs. (a) y missing from DE • Substitute p = y 0 = dy dx (so y 00 = dp ). dx Example 1.6. Solve: xy 00 + 2y 0 = 6x using reduction of order. Solution. Employing our substitutions yields dp + 2p = 6x dx dp 2 ⇒ + p = 6. dx x xy 00 + 2y 0 = 6x ⇒ x 4 This is a first-order linear DE in p, which is readily solved using an appropriate dy dy , so dx = y0 = integrating factor to give us p = 2 + C1 x−2 . But p = dx 2x + C1 x−2 . Integrating both sides with respect to x gives the solution y = x2 + Cx1 + C2 . (b) x missing from DE • Procedure: Substitute p = y 0 (so y 00 = dp dx = dp dy · dy dx =p· dp ). dy Example 1.7. Solve: yy 00 = (y 0 )2 using reduction of order. dp . Substituting these expressions into our Solution. Let p = y 0 . Then y 00 = p dy dp DE yields yp dy = p2 , which is a separable equation in p with the solution dy dy p = C1 y. But p = dx , so dx = C1 y, which is also separable, and has the Bx solution y = Ae . Exact Equations • M dx + N dy = 0, where My = Nx . • This means there exists a function F such that F = R M dx = R N dy = C. Example 1.8. Solve: (4x − y) dx + (6y − x) dy = 0. Solution. Applying the exactness test, we see that My = Nx = −1, so the equation is exact. Thus, Z Z F = M dx = (4x − y) dx = 2x2 − xy + g(y) = C and Z F = Z N dy = (6y − x) dy = 3y 2 − xy + f (x) = C. Comparing these two expressions for F , we get 2x2 − xy + 3y 2 = C. 2 Linear Equations of Higher Order Background Info Definition 2.1 (Complementary function). Let y1 , y2 , . . . , yn be n solutions of the nth-order homogeneous equation. The complementary solution is yc = c1 y1 + c2 y2 + · · · + cn yn . • The solution to a homogenous DE is its complementary solution. 5 • The solution to a nonhomogenous DE (its “general solution”) is the sum of complementary solution and any particular solution (i.e., solution that satisfies the nonhomogeneous DE). Definition 2.2 (Linear Independence/Linear Dependence). The functions f1 , f2 , . . . , fn are said to be linearly dependent on the interval I provided there exist constants c1 , c2 , . . . , cn not all zero, such that c1 f 1 + c2 f 2 + · · · cn f n = 0 on I. Otherwise, the functions f1 , f2 , . . . , fn are said to be linearly independent. Definition 2.3 (Wronskian). The Wronskian of f1 , f2 , . . . , fn is W = f1 f10 .. . (n−1) f1 f2 f20 .. . (n−1) f2 ··· ··· ··· fn fn0 .. . . (n−1) fn Theorem 2.4 (Wronskian test for linear dependence/independence). If W (f1 , f2 , . . . , fn ) = 0, then f1 , f2 , . . . , fn are linearly dependent. If W (f1 , f2 , . . . , fn ) 6= 0, then f1 , f2 , . . . , fn are linearly independent. Example 2.5. Show that y1 = e−3 , y2 = cos(2x) and y3 = sin(2x) are linearly independent using the Wronskian. Solution. e−3x cos(2x) sin(2x) W = −3e−3x −2 sin(2x) 2 cos(2x) 9e−3x −4 cos(2x) −4 sin(2x) = 26e−3x 6= 0. Methods of Solutions Homogeneous DE: Characteristic Equation • Assume the solution of the DE is of the form y = erx . Now, find r. There are three types of roots we need to consider. 1. The roots are real and distinct. If the roots of the characteristic equation, r1 , r2 , . . . , rn , are real and distinct, then the general solution of the DE is y(x) = c1 er1 x + c2 er2 x + · · · + cn ern x . 6 Example 2.6. Solve the IVP: y (3) + 3y 00 − 10y 0 = 0; y(0) = 7, y 0 (0) = 0, y 00 (0) = 70. Solution. The characteristic equation is 0 = r3 + 3r2 − 10r = r(r + 5)(r − 2). Thus, the roots are r = 0, −5, 2, and so the general solution of the DE is y(x) = c1 + c2 e−5x + c3 e2x . Solve for c1 , c2 and c3 using the ICs: c1 = 0, c2 = 2, c3 = 5. Thus, y(x) = 2e−5x + 5e2x . 2. The roots are complex and distinct. If the characteristic equation has an unrepeated pair of complex conjugate roots a ± bi (b 6= 0), then the corresponding part of the general solution is eax (c1 cos(bx) + c2 sin(bx)). Example 2.7. Solve: y 00 − 4y + 5y = 0. Solution. The roots of the characteristic equation 0 = r2 − 4r + 5 are r = 2 ± i. Thus, y(x) = e2x (c1 cos x + c2 sin x). 3. The roots are repeated. If the characteristic equation has a repeated root r of multiplicity k, then the part of the general solution corresponding to r is of the form (c1 + c2 x + · · · + ck xk−1 )erx . Example 2.8. Solve: 9y (5) − 6y (4) + y (3) = 0. Solution. The characteristic equation is 0 = 9r5 − 6r4 + r3 = r3 (3r − 1)2 . The roots are 0 (with multiplicity 3) and 31 (with multiplicity 2). Thus, the solution to the DE is y(x) = (c1 + c2 x + c3 x2 ) + (c4 + c5 x)e1/3x . Example 2.9. The roots of a characteristic equation of a certain homogeneous DE are 0, 0, 0, 0, 3, −5, −5, 2 ± 3i, 2 ± 3i. What is a general solution of this DE? Solution. The solution is y(x) = c1 + c2 x + c3 x2 + c4 x3 + c5 e3x + (c6 + c7 x)e−5x + e2x [c8 cos(3x) + c9 sin(3x)] + xe2x [c10 cos(3x) + c11 sin(3x)] 7 Method of Undetermined Coefficients • Method to find the particular solution yp of a nonhomogeneous equation with constant coefficients. • For the method of undetermined coefficients to work, the right-hand side must have finitely many linearly independent derivatives. • Procedure: 1. Find yc , the complementary solution (i.e., solution of the associated homogeneous equation) by the characteristic equation, as discussed above. 2. Find a trial solution yp by examining the right-hand side of the DE. Adjust yp to avoid linear dependence with yc , if necessary. 3. Plug yp and its derivatives into the DE. Solve for the coefficients. Example 2.10. Find a particular solution of y (3) + y 00 = 3ex + 4x2 . Solution. The steps of the solution are as follows. 1. Find yc . The characteristic equation is 0 = r3 +r2 = r2 (r +1), which has roots 0, 0, −1. Therefore, yc = c1 + c2 x + c3 e−x . 2. Find the form of the trial solution yp . By examining the right-hand side of the DE, our first try would be yp = Aex + (B + Cx + Dx2 ), where Aex is included to cover the 3ex term and B + Cx + Dx2 is included to cover 4x2 and its derivatives. However, B + Cx + Dx2 “overlaps” with terms of yc , specifically c1 and c2 x. To ensure linear independence of the terms of yc and yp , we need to multiply the term B + Cx + Dx2 of yp by x2 . Consequently, the appropriate form of yp is yp = Aex + (B + Cx + Dx2 )x2 = Aex + Bx2 + Cx3 + Dx4 . 3. Plug yp and its derivatives into the DE and solve for the unknown coefficients: 3ex + 4x2 = yp(3) + yp00 = (Aex + 6C + 24D) + (Aex + 2B + 6Cx + 12Dx2 ) = 2Aex + (2B + 6C) + (6C + 24D)x + 12Dx2 . Comparing coefficients, we see that 2A = 3, 2B +6C = 0, 6C +24D = 0, and 12D = 4. Thus, A = 23 , B = 4, C = − 34 and D = 13 . Consequently, 4 1 3 yp = ex + 4x2 − x3 + x4 . 2 3 3 8 Variation of Parameters • Method to find the particular solution yp of a nonhomogeneous equation with constant coefficients. • The right-hand side does not have to have only finitely many linearly independent derivatives for the method of variation of parameters to work. • We studied variation of Parameters formula for the nonhomogenous equation y 00 + P (x)y 0 + Q(x)y = f (x) • Procedure 1. Find solutions to the associated homogeneous equation: y1 and y2 2. Calculate W (y1 , y2 ), the Wronskian of y1 and y2 . 3. Use the following formula: Z Z y1 f (x) y2 f (x) dx y1 + dx y2 , yp = − W (y1 , y2 ) W (y1 , y2 ) Example 2.11. Find a particular solution of the DE y 00 + 3y 0 + 2y = 4ex using variation of parameters. Solution. The steps of the solution are as follows. 1. Find y1 and y2 . The characteristic equation of the associated homogeneous equation is 0 = r2 + 3r + 2 = (r + 1)(r + 2), which has roots -1 and -2. Thus, y1 = e−x and y2 = e−2x . 2. The Wronskian is given by W = e−x e−2x −e−x −2e−2x = −e−3x 3. The solution of yp is Z y1 f (x) y2 f (x) dx y1 + dx y2 yp = − W (y1 , y2 ) W (y1 , y2 ) Z −2x Z −x e · 4ex e · 4ex −x =− dx e + dx e−2x −3x −3x −e −e 4 = 2ex − ex 3 2 x = e . 3 Z 9 3 Power Series Methods • Power series methods work for variable coefficients. (The method of undetermined coefficients and the method of variation of parameters only work for constant coefficients.) • Procedure 1. Assume a power series solution of the form y = P∞ n=0 cn x n 2. Plug the power series for y and its derivatives into the DE 3. Solve for the coefficients (a) Find a recurrence relation for the coefficients (b) Use this recurrence relation to find a closed form for cn P n 4. Plug cn back into y = ∞ n=0 cn x . This is your power series solution 5. If possible, write the solution in terms of elementary functions 6. Find the radius of convergence. • Radius of Convergence P n – Given y = ∞ n=0 cn (x − a) , the radius of convergence is is at least the distance from a to the nearest singularity. – ρ := limn→∞ cn cn+1 ∗ ρ = 0 ⇒ series diverges for all nonzero x ∗ ρ = ∞ ⇒ series converges for all x ∗ 0 < ρ < ∞ ⇒ series converges for |x| < ρ; diverges for |x| > ρ Example 3.1. Solve the DE: (x − 3)y 0 + 2y = 0 and find its radius of convergence. P∞ P n 0 x−1 . Substituting these into the DE Solution. Let y = ∞ n=0 cn x . Then y = n=1 cn nx 10 gives: 0 = (x − 3)y 0 + 2y = xy 0 − 3y 0 + 2y ∞ ∞ ∞ X X X =x cn nxn−1 − 3 cn nxx−1 + 2 cn x n n=1 = = ∞ X n=1 ∞ X n=1 cn nxn − 3 n cn nx − 3 n=0 ∞ X n=1 ∞ X cn nxn−1 + 2 n=0 ∞ X cn x n n=0 n cn+1 (n + 1)x + 2 n=0 ∞ X cn x n n=0 ∞ X = [cn n − 3cn+1 (n + 1) + 2cn ]xn n=0 ∞ X = [(n + 2)cn − 3cn+1 (n + 1)]xn n=0 By the identity principle, (n + 2)cn − 3cn+1 (n + 1) = 0, from which we get cn+1 = 1 n+2 · · cn , n ≥ 0. 3 n+1 Reindexing for n ≥ 1, we see that cn = 1 n+1 · · cn−1 , n ≥ 1. 3 n c1 = 32 c0 c2 = 3 c 3·2 1 = 3 3·2 2 c 3 0 c3 = 4 c 3·3 2 = 4 3·3 3 c 32 0 .. . cn = n+1 c, 3n 0 = = 3 c 32 0 4 c. 33 0 n≥1 11 Therefore, y= ∞ X cn x n n=0 ∞ X n+1 = c0 x n n 3 n=0 ! ∞ X n+1 xn . = c0 n 3 n=0 From the recurrence relation, cn cn+1 =3 n+1 n+2 . Therefore, the radius of convergence is 1+ ρ = 3 lim n→∞ 1 + 1 n 2 n = 3. Consequently, the series converges if |x| < 3 and diverges if |x| > 3. 12