Assignment # 3, Math 370, Fall 2018 SOLUTIONS:

Assignment # 3, Math 370, Fall 2018 SOLUTIONS: Problem 1: Solve the equations: (a) y (1 + x)e x y 2 = xy, (i) y(0) = 1, (ii) y(0) = 0. On what intervals are the solution of the IVP defined? (b) 2y + y cosx = cosx(1 + sin x)y 1. Solution: (a) We can rewrite y + xy = (1 + x)e x y 2. This is Bernoulli with α = 2. Substitution z = y 1 α = 1 y. Then z = 1 y 2 y. We multiply the equation by 1 y 2 to obtain 1 y 2y x 1 y = (1 + x)e x or z xz = (1 + x)e x. We have P(x) = x so integrating factor is e 1 2 x2 and we obtain e 1 2 x2 z xe 1 2 x2 z = (1 + x)e x 1 2 x2 or (e 1 2 x2 z) = (1 + x)e x 1 2 x2. e 1 2 x2 z = e x 1 2 x2 + C z = e x + Ce 1 2 x2. We obtain 1 y = and the special solution y 0. e x + Ce 1 2 x2 IVP: (i) We need 1 = 1 so C = 0. Solution: y 1+C 1 = 1 = e x. e x (ii) Special solution y 0 satisfies this initial condition. Both solutions defined on (, + ). (b) 2y + y cos x = cosx(1 + sinx)y 1. This is Bernoulli with α = 1. Substitution z = y 1 α = y 2. Then z = 2yy. We multiply the equation by y to obtain 2yy + y 2 cos x = cos x(1 + sinx) or z + cos xz = cos x(1 + sinx). We have P(x) = cos x so integrating factor is e sinx and we obtain e sinx z + cos xe sinx z = cos x(1 + sinx)e sinx or (e sinx z) = cos x(1 + sinx)e sinx. e sinx z = We obtain cos x(1+sinx)e sinx dx+c = {t = sinx} = (1+t)e t dt+c = te t +C = sinxe sinx +C. z = sinx + Ce sinx. y = ± sinx + Ce sinx and the special solution y 0. 1

2 Problem 2: Solve the equation: (a) ( y x 2 cos y x) dx ( 1 x cos y x + 2y) dy = 0, y(1) = π. On what interval is the solution of IVP defined? Find such that the equation is exact and solve it (b) (3x 2 + y 2 )dx + (8xy + e y )dy = 0, y(1) = 1. Solution: (a) We will try if it is exact: M = y x 2 cos y x and N = ( 1 x cos y x + 2y) so M y = 1 x 2 cos y x y x 2 1 x sin y x, N x = 1 x cos y 2 x 1 y x x sin y 2 x and the equation is exact. We are looing for a function f such that { (1) f x = y cos y ; x 2 x (2) f y = 1 cos y 2y. x x Then, integrating second equation in y we obtain f = x 1 x sin y x y2 + C(x) = sin y x y2 + C(x), which implies f x = y x cos y 2 x + C (x). Comparing with (1) we obtain C (x) = 0 so C(x) = C and The solution (implicit solution) is f = sin y x y2 + C. sin y x y2 = K. (b) Let M = 3x 2 + y 2 and N = 8xy + e y. Then, M y = 2y, N x = 8y so the equation is exact for = 4. We are looing for a function f such that { (1) f x = 3x 2 + 4y 2 ; (2) f y = 8xy + e y. Then, integrating equation (1) in x we obtain f = x 3 + 4xy 2 + C(y), which implies f y = 8xy + C (y).

3 Comparing with (2) we obtain C (y) = e y so C(y) = e y + C and The solution (implicit solution) is Problem 3: f = x 3 + 4xy 2 + e y + C. x 3 + 4xy 2 + e y = K. Consider the equation dy dx = 4 x 1 2 x y + y2. It is an example of Riccati equation. (a) Chec that y 1 = 2 is a solution. x (b) Substitute y = y 1 + u and obtain a Bernoulli equation. (c) Solve the equation. Solution: (a) We have 2 = 4 1 2 + 4 which shows that y x 2 x 2 x x x 2 1 is a solution. (b) We substitute and obtain: 2 x + 2 u = 4 x 1 2 2 x x 1 x u + 4 x + 22 2 x u + u2, which reduces to a Bernoulli equation ( ) u 3 x u = u2. We have α = 2 so we mae substitution z = u 1 = 1. We have u so we multiply ( ) by 1 u 2 and obtain ( ) z = 1 u 2u, 1 + 3 1 u 2u x u = 1 or z + 3 x z = 1. This is a linear equation. We have P(x) = 3 x so the integrating factor is µ = er Pdx = e 3ln x +C = Dx 3 but we sip D as it does not change anything. We obtain x 3 z + 3x 2 z = x 3 or (x 3 z) = x 3 x 3 z = 1 4 x4 + C. This gives We also have a special solution z = 1 4 x + C x 3. u = 1 1 4 x + C x 3. u 0.

4 and special solution y = 2 x + 1 1 4 x + C x 3, y = 2 x. Problem 4: Find N(x, y) such that the equation ( y 2 + y cos 2 x is exact. Solve the equation. Solution: We have M y = 2y + 1 cos 2 x For example ) dx + N(x, y)dy = 0, so N can be any function satisfying N x = 2y + 1 cos 2 x, N(x, y) = 2xy + tan x. Now, we solve the equation ( y 2 + y ) dx + (2xy + tan x)dy = 0. cos 2 x We now that it is exact. We are looing for a function f such that { (1) f = y 2 + y ; x cos 2 x (2) f = 2xy + tan x. y Then, integrating equation (1) in x we obtain f = xy 2 + y tanx + C(y), which implies f y = 2xy + tanx + C (y). Comparing with (2) we obtain C (y) = 0 so C(y) = C and The solution (implicit solution) is f = xy 2 + y tan x + C. xy 2 + y tan x = K. Problem 5: A large tan is filled with 500 gal of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tan at the rate of 5 gal/mi. The well-mixed solution is pumped out at the rate 10 gal/min. What is the concentration of salt in the tan when it is half full?

5 Solution: Let S(t) denote the amount of salt in the tan. We have ds dt = IN OUT = 2 5 S 10 500 5t, since the amount of water in the tan changes by 5 gal/min. We solve the equation ds dt + S 10 500 5t = 10. We have P(t) = 10 so P(t)dt = 10 ln 100 t and the integrating factor is µ = 500 5t 5 (100 t) 2. We obtain the equation (100 t) 2 S + or (100 t) 2 S 10 = 10(100 t) 2 or ((100 t) 2 S) = 10(100 t) 2. 500 5t (100 t) 2 S = 10(100 t) 1 + C, S = 10(100 t) + C(100 t) 2 = 1000 10t + C(100 t) 2. We obtain C from the condition S(0) = 0: 0 = 1000 + C(100) 2 or C = 1/10 and S = 1000 10t 1 10 (100 t)2. Since the tan will be half full after 50 min we need S(50): S(50) = 1000 10 50 1 10 (50)2 = 495 lb. the concentration then is 495/250 1.98 lb/gal. Problem 6: Air Resistance: Suppose that a cannonball weighing 16 pounds is shot vertically with initial velocity v 0 = 300 ft/s. Answer the question: How high does the cannonball go under the assumptions (a),(b)? (a) Ignore the air resistance. Tae g = 32 ft/s 2. (b) Assume that the air resistence is proportional to instantaneous velocity, i.e., F r = v, with = 0.0025. Solution: (a) Let us orient the y axis upward and assume that the ball starts at hight 0. The only force acting is gravity so ma = mg. We put minus sign since gravity acts against the movement of the ball. This means: dv = g dt v = gt + C. Since v(0) = 300 we obtain C = 300 and v = gt + 300.

6 The ball stops going up when v = 0. We have 0 = gt + 300 or t = 300 = 300 g 32 Apparently this time is independent of the mass of the ball. Now, the distance: we have s = vdt = ( gt + 300)dt = 1 2 gt2 + 300t + D. The initial condition s(0) = 0 gives D = 0 so we have s = 1 2 gt2 + 300t and s(9.375) = 1 2 32(9.375)2 + 300(9.375) 1406.25 ft. (b) Now, the equation is ma = mg v, where = 0.0025. We have dv dt = g m v or dv dt + m v = g. This is a linear equation. P = /m so µ = e m t and we obtain and e m tdv dt + m e m t v = ge m t or (e m t v) = ge m t. e m t v = g m e m t + C, v = mg + Ce Initial condition v(0) = 300 implies C = 300 + mg The velocity is 0 when which gives v = mg + ( mg 300 + mg so 300 + mg m t. = e m t ) e m t. 9.375 sec. t = m ( 300 + mg ) ln mg 9.16. We needed the mass: w = mg so m = 16/32 = 1/2. We can see that with air resistance the ball will stop to ascend earlier. Now, the maximal height. We have ( s = vdt = mg ( + 300 + mg ) )e m t dt = mg ( t 300 + mg ) m e m t + D. The initial condition s(0) = 0 gives ( D = 300 + mg ) m,

7 so We have s = mg ( t + 300 + mg ) m s(9.16) = 1363.79 ft, (1 e m t ). which is lower than before. Problem 7: Leaing Cylindrical Tan: A tan in the form of a right-circular cylinder standing on end is leaing water through a circular hole in its bottom. When friction and contraction of water at the hole are ignored, the height h of water in the tan is described by dh dt = A h 2gh, where and A h are the cross-sectional areas of the water and the hole, respectively. (a) Solve the DE if the initial height of the water is H. (b) Suppose the tan is 20 feet high and has radius 2 feet and the circular hole has radius 1 inch. If the tan is initially full, how long will it tae to empty? Use g = 32ft/s 2. Solution: (a) We solve the equation dh dt = A h 2gh. It is separable so 1 dh = A h dt 1 A h 2gh = t + C or h = g ( A ) 2 h t + C. 2gh g 2 If h(0) = H then we have H = g 2 C2 2H and C =. We have g h = g 2 ( A h t + ) 2 2H. g (b) The tan is empty when h = 0 so we need A h 2H t + = 0 or g t = 2HAw gah. We have H = 20, = π2 2 and A h = π(1/12) 2 in feet and feet 2. We use g = 32ft/s 2. Then t = 2 10 4π 4 2 π(1/144) = 20 144 643.99 sec.

8 Problem 8: Conical Tan: When friction and contraction of the water at the hole are taen into account, the model in Problem 7 becomes dh dt = ca h 2gh, where 0 < c < 1. A tan in the form of a rightcircular cone standing on end, vertex down, is leaing water through a circular hole in its bottom. Suppose the tan has a vertex angle of 60 and the circular hole has radius 2 inches. Determine the differential equation governing the height h of water. If the height of the water is initially 9 feet, how long will it tae the tan to empty? Tae the friction/contraction coefficient to be c = 0.6 and g = 32ft/s 2. Solution: Moved to the next assignment