3. Solve: x' = (7x - 2)t*, x(0) = 1 [Your answer should be solved for x] This equation is separable: dx dt = (7x - 2)t4 1 7x - 2 -dx = t*dt 1 7x - 2' 7dx = t+dt = In|7x - 21 = = +5 + q, In|7x - 2| = =t's 17x - 21 = est +62 = ests . etz = Casts 7x - 2 = tegests _ crests x = - CAest + 2 = cests + 2 7 Applying the initial condition: x(0) = 1 = ce+ = =1= c= So the solution is: 5 7 + : JIN X= = 4. Use the linear method to solve: xy' - 4x y = 2x2; y(0) = 2 Rewriting in the "standard" form: y' - 4xy = 2x Then the integrating factor is r(x) = e-4xdx = e-2x. Multiplying through yields: e-2x y' - 4xe-2x y = 2xe-2x2 y [e zx* y] = 2xe-2x2 e-2xly = 2xe-2x2 dx e-2x y = = =e-2x2 + c 1 y = - 7+ Ce2x2 Applying the initial condition: y(0) = 2= -=+c.e' = 2= c= So the solution is: 1 y = e2x2 2 Page 3 of 71. Given the equation xzy" + 5xy' -12y = 0. [Note that we have not yet learned how to solve this equation in our course.] a. Verify that y = Cix2 + C2x-6 is a solution. Assuming y = Cix2 + C2x 6, then y' = 2cix - 602x-7, and y" = 2c1 + 42czx-8. Substituting into the given differential equation yields: x2 [2c1 + 42c2x-8] + 5x[2c,x - 6C2x-7] - 12[c,x2 + C2x-6] = 0 2c1x2 + 42c2x-6 + 10c,x2 - 30c2x-6 - 12c,x2 -12czx-6 = 0 [2 + 10 - 12]c1x2 + [42 -30 - 12]C2x-6 = 0 [0]c,x2 + [0]C2x-6 = 0 0 =0 So, y = Cix2 + C2x 6 is a solution to the differential equation. b. Solve the initial-value problem: xly" + 5xy' -12y = 0; y(1) = 5, y'(1) = -6 Since y = C1x2 + C2x-6 is a solution to the differential equation, then applying the initial conditions: y(1) = 5 = c, (1)2 + C2(1)-6 = 5 = q1 + C2 = 5 y'(1) = -6= 2c, (1) - 602(1)-7=-6= 201 - 602 = -6 Solving simultaneously for c, and c2 yields c1 = 3, C2 = 2. So the specific solution for the initial-value problem is y = 3x2 + 3x-6 2. Solve: y' = 2x3 + e-5x + sin 3x ; y(0) = 2 dx ay = 2x3 + e-5x + sin 3x dy = (2x3 + e-5x + sin 3x)dx dy = (2x3 + e-5x + sin 3x) dx 1 e-5x _ 1 5 3 cos 3x + C2 1 y = =x4 - se 5x_ 1 cos 3x + c Applying the initial condition: y(0) = 2 2 = = (0)4 - 1 ze-5.0 _ 1 5 - cos (3 . 0) + c 1 1 2= 0 + c c = 2+: 1 38 + 5 15 So the specific solution is y = = x e-5x cos 3x + 15 Page 2 of 7Test Exercise 3. Solve: x' = (5x + 3)t3, x(0) = 1 [Your answer should be solved for x] Test Exercise 4. Use the linear method to solve: xy' - 6x2y = 3x2; y(0) = 3 Page 3 of 7Test Exercise 1. Given the equation x2y" - 4xy' - 14y = 0. [Note that we have not yet learned how to solve this equation in our course.] a. Verify that y = cix 2 + C2x is a solution. b. Solve the initial-value problem: x2y" - 4xy' -14y = 0 ; y(1) = 5, y'(1) = -6 Test Exercise 2. Solve: y' = 3x5 + e 4* + cos 5x ; y(0) = 3 Page 2 of 7