Practice: 3.1-3.2, Systems of Differential Equations

Practice: 3.1-3.2, Systems of Differential Equations 1. For each second order differential equation below, write the corresponding system of first order equations. Lastly, write the system in matrix-vector form. (a) y 00 + 2y 0 − 3y = 0 (c) y 00 − 9y = 0 (b) y 00 + 4y 0 + 4y = 0 (d) y 00 − 2y 0 + 2y = 0 2. For extra practice, go back and give the general solution to each second order DE (using methods from Chapter 3 of our textbook). 3. For each system of first order, convert to a corresponding second order differential equation. (a) x0 = 3x − 2y y 0 = 2x − 2y (b) x0 = −2x + y y 0 = x − 2y (c) x0 = x + y y 0 = 4x − 2y 4. For extra practice, solve each of the second order DEs from the previous exercise using techniques from Chapter 3 of our text. 5. Consider the system below: x0 = −3x + y y 0 = −2y Solve this system by recognizing that we can solve for y directly, then substitute this into the DE for x and solve it as a first order linear DE. 6. Each system below is nonlinear. Solve each by first writing the system as dy/dx. (a) x0 = y(1 + x3 ) y 0 = x2 (b) x0 = 4 + y 3 y 0 = 4x − x3 (c) x0 = 2x2 y + 2x y 0 = −(2xy 2 + 2y) 7. Convert the third order (nonlinear) differential equation into a system of first order equations. y 000 − y 00 + y 0 y = t2 1 SOLUTIONS 1. For each second order differential equation below, write the corresponding system of first order equations. (a) y 00 + 2y 0 − 3y = 0 SOLUTION: Let u = y, v = y 0 . Then notice that v 0 = y 00 = 3y − 2y 0 = 3u − 2v. The full system is then:    0   0 1 u u0 = v u = ⇒ 0 0 3 −2 v v v = 3u − 2v NOTE: Always write your variables in the same order. Since we started with u, v, then u0 came first, then v 0 , etc. (b) y 00 + 4y 0 + 4y = 0 SOLUTION: Let u = y, v = y 0 . Then v 0 = y 00 = −4y − 4y 0 = −4u − 4v, and    0   0 1 u u0 = v u = ⇒ 0 0 v −4 −4 v v = −4u − 4v (c) y 00 + 9y = 0 SOLUTION: Let u = y, v = y 0 . Then v 0 = y 00 = −9y = −9u, and    0   0 1 u u0 = v u = ⇒ 0 0 v −9 0 v v = −9u (d) y 00 − 2y 0 + 2y = 0 SOLUTION: Let u = y, v = y 0 . Then v 0 = y 00 = 2y 0 − 2y, and    0   0 1 u u0 = v u = ⇒ 0 0 v −2 2 v v = −2u + 2v 2. For extra practice, go back and give the general solution to each second order DE (using methods from Chapter 3 of our textbook). SOLUTIONS: (a) y 00 + 2y 0 − 3y = 0 The characteristic equation is r2 + 2r − 3 = 0, or (r + 3)(r − 1) = 0, so r = 1, −3. Therefore, the general solution is y(t) = C1 et + C2 e−3t (b) y 00 + 4y 0 + 4y = 0 The characteristic equation is r2 + 4r + 4 = 0, or (r + 2)2 = 0, so r = −2, −2. Therefore, the general solution is y(t) = e−2t (C1 + C2 t) 2 (c) y 00 + 9y = 0 The characteristic equation is r2 + 9 = 0, or r = ±3i, y(t) = C1 cos(3t) + C2 sin(3t) (d) y − 2y 0 + 2y = 0 The characteristic equation is r2 −2r+2 = 0, or r = 1±i, so r = 1, −3. Therefore, the general solution is y(t) = et (C1 cos(t) + C2 sin(t)) 3. For each system of first order, convert to a corresponding second order differential equation. (a) x0 = 3x − 2y y 0 = 2x − 2y SOLUTION: Using the first equation to solve for y, y = −(1/2)(x0 − 3x). Putting this into the second equation: 0  0   0 x − 3x x − 3x = 2x − 2 − − 2 2 Clean up by multiplying both sides by −2: x00 − 3x0 = −4x − 2(x0 − 3x) (b) x00 − x0 − 2x = 0 x0 = −2x + y y 0 = x − 2y SOLUTION: Using the first equation to solve for y, y = x0 + 2x. Putting this into the second equation: 0 (x0 + 2x) = x − 2 (x0 + 2x) (c) ⇒ ⇒ x00 + 4x0 + 3x = 0 x0 = x + y y 0 = 4x − 2y SOLUTION: Similar to the others, y = x0 − x from equation 1, so equation 2 becomes: x00 − x0 = 4x − 2(x0 − x) ⇒ x00 + x0 − 6x = 0 4. For extra practice, solve each of the second order DEs from the previous exercise using techniques from Chapter 3 of our text. 3 (a) x(t) = C1 e2t + C2 e−t (b) x(t) = C1 e−t + C2 e−3t (c) x(t) = C1 e2t + C2 e−3t 5. Consider the system below: x0 = −3x + y y 0 = −2y Solve this system by recognizing that we can solve for y directly, then substitute this into the DE for x and solve it as a first order linear DE. SOLUTION: Looking at y 0 = −2y, y(t) = C1 e−2t . Putting that into the equation for x0 , we get: x0 + 3x = C1 e−2t The integrating factor is e3t , so that (xe3t )0 = C1 et ⇒ xe−3t = C1 et + C2 As a parametric system, we have:  ⇒ C1 e−2t + C2 e−3t C1 e−2t x(t) = C1 e−2t + C2 e−3t  We would note that the C1 for y is the SAME as the C1 for the x. 6. Each system below is nonlinear. Solve each by first writing the system as dy/dx. (a) x0 = y(1 + x3 ) y 0 = x2 SOLUTION: This becomes separable. Z Z 2 dy x2 x dx = ⇒ y dy = 3 dx y(1 + x ) 1 + x3 ⇒ 1 2 1 y = ln(1 + x3 ) + C 2 3 At this point, we’ll leave it since we won’t know if we need the positive or negative root. x0 = 4 + y 3 (b) 0 y = 4x − x3 SOLUTION: This becomes separable: Z Z 4x − x3 1 1 dy 3 ⇒ 4+y dy = 4x−x3 dx ⇒ 4y+ y 4 = 2x2 − x4 +C = 3 dx 4+y 4 4 We’ll leave this in implicit form. (c) x0 = 2x2 y + 2x y 0 = −(2xy 2 + 2y) 4 SOLUTION: This becomes exact (don’t cancel anything!) dy 2xy 2 + 2y =− 2 dx 2x y + 2x (2xy 2 + 2y) + (2x2 y + 2x) ⇒ dy =0 dx This is exact, with My = Nx = 4xy + 2. The solution is x2 y 2 + 2xy = C 7. Convert the third order (nonlinear) differential equation into a system of first order equations. y 000 − y 00 + y 0 y = t2 SOLUTION: Let x1 = y, x2 = y 0 , and x3 = y 00 . Then: x01 = x2 x02 = x3 x03 = x1 x2 + x3 + t2 5