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First-order Equations 1. Determine, for each equation, (a) the order. (b) whether the equation is linear. (c) whether the equation is separable. (d) whether the equation is homogeneous. • y ′ = −xy 1st order. Linear and Separable. • y ′ + xy = x3 1st order. Linear. • y ′ yx + x2 = y 2 1st order. Homogeneous. x −e • y ′ = e 3+4y 1st order. Separable. −x • y ′′ + cos(t)y ′ + 2t y = tet 2nd order. Linear. 2 +3y • y ′′′ = x 2xy 3rd order. 2 • y ′′ + xy ′ = x 2nd order. Linear and Separable. 2 +4x+2 • y ′ = 3x2(y−1) 1st order. Separable. 2. Sketch the slope field and describe the long-term behavior of the solutions. (a) y ′ = −3y + 5 (stable) equilibrium: y = 5/3 Long term behavior: limx→∞ y = (b) y ′ = 5 3 −y x (stable) equilibrium: y = 0 Long term behavior: limx→∞ y = 0 1 (c) y ′ = −3y 2 + 21y − 30 (stable) equilibrium: y = 5 (unstable) equilibrium: y = 2 −∞, 2, Long term behavior: limx→∞ y = 5, if y(0) < 2 if y(0) = 2 if y(0) > 2 3. Solve the IVP and determine the domain of your solution: 12 − xy ′ = 3y (a) y(1) = 3 xy ′ + 3y = 12 y′ + µ(x) = e R x3 y = 4x3 + C 3 12 y= x x 3 x dx 3=4+C C = −1 = e3 ln x = x3 y =4− (x3 y)′ = x3 y ′ + 3x2 y = 12x2 1 x3 Domain: 0 < x < ∞. (b) ( 2 y y ′ = yx2 −1 y(3) = e (y 2 − 1)dy = x2 dx y Z Z 1 (y − )dy = x2 dx y 1 1 2 e − 1 = (33 ) + C 2 3 1 2 C = e − 10 2 1 2 1 y − ln |y| = x3 + C 2 3 1 2 2y 63 3 2 − e 6= x3 2 2 r 3 3 x 6= (21 − e2 ) < 3 2 q Domain: 3 32 (21 − e2 ) < x < ∞ Whew! Domain: y 2 − 1 6= 0 y 6= ±1 1 1 1 6= x3 + e2 − 10 2 3 2 (c) y ′ − 4tan(4t)y = et y(0) = 0 µ(t) = e − ln |y| = 31 x3 + 21 e2 − 10 R −4 tan(4t)dt = eln(cos(4t)) = cos(4t) (cos(4t)y)′ = cos(4t)y ′ − 4 sin(4t)y = et cos(4t) 2 cos(4t)y = Z et cos(4t)dt = Re Z e(1+4i)t dt = Re 1 e(1+4i)t dt + C 1 + 4i 1 t 4 1 − 4i (1+4i)t e dt + C = e cos(4t) + et sin(4t) + C cos(4t)y = Re 17 17 17 y= y= 1 t 17 e + 4 t 17 e 1 t 4 e + et tan(4t) + C 17 17 1 0= +C 17 1 C =− 17 tan(4t) − 3 1 17 Domain: − π8 < x < π 8