20D - Homework Assignment 1

Transcription

1 0D - Homework Assignment Brian Bowers (TA for Hui Sun) MATH 0D Homework Assignment October 7, 0. #,,,4,6 Solve the given differential equation. () y = x /y () y = x /y( + x ) () y + y sin x = 0 (4) y = (x )/( + y) (6) xy = ( y ) / All of the equations in this set are separable, so our goal should be to get the terms involving y on the left side and the terms involving x on the right side. I will rewrite y as /dx for these problems. () We separate and then integrate: dx = x /y = y = x dx [separating variables] = y = x + C [integrating] x = y = ± + C [solving for y] () We separate and then integrate: dx = x /y( + x ) = y = x dx [separating variables] + x = y = ln + x + C [integrating] = y = ± ln + x + C [solving for y] () We separate and then integrate: dx = y sin x = = sin xdx [separating variables] y = = cos x + C [integrating] y = y = [solving for y] cos x + C

2 (4) We separate and then integrate: dx = (x )/( + y) = ( + y) = (x )dx [separating variables] = y + y = x x + C [integrating] = (y + /) = x x + C [completing the square] = y = ± x x + C / [solving for y] (6) We separate and then integrate: x dx = ( y ) / = ( y ) / = dx [separating variables] x = sin (y) = ln x + C [integrating] = y = sin(ln x + C) [solving for y]

3 . #9a, 7a, 9a Find the solution of the given initial value problem in explicit form. (9a) y = ( x)y, y(0) = /6 (7a) y = (x e x )/(y 5), y(0) = (9a) sin xdx + cos y = 0, y(π/) = π/ First, we should clarify what it means to find an explicit solution - an explicit solution is a solution of the form y = f(x) (as opposed to something like y + y = x + ). Once again, for each problem, I will write y as /dx for ease of separation. (9a) We separate and then integrate: = ( x)y dx = = ( x)dx [separating variables] y = y = x x + C [integrating] = y = x x [solving for y] + C Next, we plug in our initial condition: y(0) = (0) (0) + C = C = 6, which we can solve to get that C = 6. Thus, plugging back into our solution for y, we get (7a) We separate and then integrate: y = x x + 6. dx = (x e x )/(y 5) = (y 5) = (x e x )dx [separating variables] = y 5y = x e x + C [integrating] ( = y ) 5 = x e x + C [completing the square] = y = ± x e x + C + 5 [solving for y] Next, we plug in our initial condition: y(0) = ± (0) e (0) + C + 5 = ± C + 5 =. So, we find that ± C =. But the square root of a number is always positive, so we choose the negative sign and get C =, which we can solve to find that C = 4. Thus, plugging back into our solution for y, we get y = x e x Note that our final solution only involves the negative of the square root since this is what we found from our initial conditions.

4 (9a) We separate and then integrate: sin xdx + cos y = 0 = cos y = sin xdx [separating variables] = sin y = cos x + C [integrating] Next, we plug in our initial condition (which says that when x = π/, we have y = π/): sin((π/)) = cos((π/)) + C = sin(π) = cos(π) + C = 0 = + C = C = Next, recall the following facts: Thus, we get the solution sin(x) = sin(π x) and sin(x) = sin(x + πk) for any integer k. sin(y) = cos(x) + = sin(y) = cos(x) + ( ) sin cos(x)+ + kπ = y = or ( ) π sin cos(x)+ + kπ If we plug our initial condition into the first formula, we get ) + kπ π sin = π π sin = ( cos((π/))+ ( cos((π/))+ = sin (0) + kπ = kπ, but this solves to k = /, which is not an integer. So, we consider the second formula: ) + kπ = π sin (0) + kπ = π + kπ, which solves as k = 0. Thus, our final explicit formula should be the second formula with k = 0, namely ( ) π sin cos(x)+ y =. 4

5 . #0(abcde) Consider the equation (a) Show that equation (??) can be rewritten as thus Equation (??) is homogeneous. dx = y 4x x y. () dx = (y/x) 4 (y/x) ; () (b) Introduce a new dependent variable v so that v = y/x or y = xv(x). Express /dx in terms of x, v, and dv/dx. (c) Replace y and /dx in Equation (??) by the expressions from part (b) that involve v and dv/dx. Show that the resulting differential equation is or. Observe that Equation (??) is separable v + x dv dx = v 4 v, x dv dx = v 4 v () (d) Solve Equation (??), obtaining v implicitly in terms of x. (e) Find the solution of Equation (??) by replacing v by y/x in the solution in part (d). First, I think it s worth clarifying what homogeneous means in this context. We are using the definition given on pages 49-50, namely If the right side of the equation /dx = f(x, y) can be expressed as a function of the ratio y/x only, then the equation is said to be homogeneous. Now let s work through the problem: (a) In order to rewrite an equation, we want to modify it in such a way that the content remains unchanged. In this case, we achieve this by multiplying the right side of the equation by in a special way. dx = y 4x x y ( ) /x y 4x dx = [multiplying by ] /x x y dx = (y/x) 4 (y/x) [simplifying] Thus, since the right side of the equation is expressed solely in terms of y/x, we conclude that the equation is homogeneous. (b) Define v = y/x. If we solve for y, we get y = xv. Differentiating with respect to x, we use the product rule to get /dx = v + x dv dx. 5

6 (c) Next, we plug substitute y/x = v and /dx = v + x dv dx : v + x dv dx = v 4 v = x dv dx = v 4 v v [rearranging] = x dv dx = v 4 v v v [getting a common denominator] v = x dv dx = v 4 [combining terms] v We could rearrange terms to write this as which means that this equation is separable. dv( v) v 4 = dx x, (4) (d) Now we solve Equation (??) using the separated version, Equation (??). In order to integrate the left side of Equation (??), we will need to use partial fractional decomposition. We need to find A and B such that v v 4 = A v + B A(v + ) + B(v ) = v + v = 4 (A + B)v + (A B) v. 4 Thus, comparing the terms with v we see that = A + B, and comparing the constant terms, we see that = A B. The first equation tells us that A = B. Pluggin this into the second equation we get ( B) B = 4B =, which we can solve to get B = /4. Substitution this back into our first equation, we get = A /4, which we can solve to get A = /4. Thus, we can rewrite Equation (??) as ( /4 v + /4 ) dv = dx v + x = 4 ln v ln v + = ln x + C [integrating] 4 (e) Last, we substitute v = y/x everywhere in our solution. We get 4 ln (y/x) ln (y/x) + = ln x + C. 4 This is technically a correct solution to the differential equation and we could end the problem here, but we have to do a bit of work to make this a simpler form (specifically a simpler form that matches the answer given in the back of the book.) If you re interested to see how to get the nicer form, read the 6

7 following steps. If not, feel free to stop the problem here. = 4 ln y x x 4 ln y + x x = ln x + C [common denominator] /4 /4 = ln y x x + ln y + x x = ln x + C [properties of logs] = e y x ln x ln y x /4 +ln y+x x /4 = e ln x +C [exponentiating] = e x /4 y+x e ln x /4 = e ln x e C [properties of exponents] /4 = y x /4 y + x x x = C x [simplifying, renaming C] ( ) /4 ( ) /4 = y x /4 y + x /4 = C x [factoring out] x x = y x /4 y + x /4 x = C x [regrouping] = y x /4 y + x /4 = C [dividing by x ] = y x y + x = C [raising both sides to the 4 power, renaming C]. #b The method outlined in Problem 0 can be used for any homogeneous equation. That is, the substitution y = xv(x) transforms a homogeneous equation into a separable equation. The latter equation can be solved by direct integration, and the replacing v by y/x gives the solution to the original equation. Solve the differential equation dx = x + y. xy We will follow the model of problem 0. (a) In order to express the right side in terms of (y/x), we need to multiply the right side by in a particular way, specifically: dx = x + y ( ) /x xy /x [multiplying by ] = + y /x y/x = + (y/x) (y/x) (b) As in problem 0, we see that /dx = v + x dv dx. 7

8 (c) Now we replace (y/x) and /dx and rearrange to make it separable: dx = + (y/x) (y/x) = v + x dv dx = + v v = x dv dx = + v v [subtracting] v = x dv dx = + v v v v = x dv dx = + v [combining terms] v = v + v dv = dx x [rearranging] [common denominator] (d) Now we integrate: v + v dv = dx x = ln( + v ) = ln x + C [integrating] = e ln(+v) = e ln x +C [exponentiating] = e ln(+v) = e ln x e C [properties of exponents] = + v = C x [simplifying, renaming C] (e) Last, we substitute v = y/x and simplify: + v = C x = + (y/x) = C x = + y x = C x = x + y = C x x = x + y = C x x. # (abridged) The half-life of carbon-4 is approximately 570 years. If Q(t) is the amount of carbon-4 at time t and Q 0 is the original amount, then the ratio Q(t)/Q 0 ca be determined, as long as this quantity is not too small. (a) Assuming that Q satisfies the differential equation Q = rq, determine the decay constant r for carbon-4. (b) Find an expression for Q(t) at any time t, if Q(0) = Q 0. (c) Suppose that certain remains are discovered in which the currend residual amount of carbon-4 is 0% of the original amount. Determine the age of these remains. 8

9 (a) Let s solve the differential equation first and see if we can answer the question. In other words, we ll be doing part (b) first. dq/dt = rq = dq/q = rdt [rearranging] = ln Q = rt + C [integrating] = Q = e rt+c [exponentiating] = Q = Ce rt [renaming C] Next, we use the initial condition, namely Q(0) = Q 0. Plugging into our equation, we get Q(0) = Ce r(0) = C = Q 0, which tells us that C = Q 0. Thus, the solution to the differential equation is Q = Q 0 e rt. Now let s use the half-life given in the problem statement. It tells us that after 570 years, the remaining amount will be half of the original amount. Written as an equation, this tells us that Q(570) = Q 0 /. Let s use this to find r. We get Q(570) = Q 0 e r(570) = Q 0 / = e r(570) = / [dividing by Q 0 ] = r(570) = ln(/) [taking ln] = r = ln(/)/570 = ln()/ (b) Looking back at part (a), we see that we basically alrea found this. We take our equation and plug in the value of r that we just found. Q = Q 0 e t (c) If the current amount is 0% of the original, then we can write this in equation form as Q(t) = Q 0 e t =.Q 0 = e t =. [dividing by Q 0 ] = t = ln(.) [taking the ln of both sides] = t = ln(.)/ [solving] 04 years [simplifying]. #5 Suppose that a certain population satisfies the initial value problem /dt = r(t)y k, y(0) = y 0, where the growth rate r(t) is given by r(t) = ( + sin t)/5 and k represents the rate of predation. (a) Suppose that k = /5. Plot y versus t for several values of y 0 between / and. (b) Estimate the critical initial population y c below which the population will become extinct. (c) Choose other values of k and find the corresponding y c for each one. (d) Use the data you have found in parts (b) and (c) to plot y c versus k. 9

10 To complete this problem, we must use differential equation graphing software of some variety. By trial and error, we can find the answers by stuing the graphs. (a) [Use software to find this step - too long to show here.] (b) By stuing the graphs from step (a), we can find that y c 5/6. (c) If we choose different values of k, we find that y c 5k/6. (d) A graph of y c versus k would look like the following:. #7a (abridged) Heat transfer from a bo to its surroundings is described by the differential equation du dt = α(u4 T 4 ), where u(t) is the absolute temperature oft he bo at time t, T is the absolute temperature of the surroundings, and α is a constant depending on the physical parameters of the bo. However, if u is much larger thatn T, then solutions of the above equation are well approximated by soutions of the simpler equation du dt = αu4. Suppose that a bo with initial temperature 000 K is surrounded by a medium with temperature 00 K and that α =.0 0 K /s. Determine the temperature of the bo at any time by solving the second equation. Putting the above conditions into equation form, we get u(0) = 000, T = 00, and α =.0 0 K /s. As it turns out, we don t even need the information about T, so we plug in α and solve: du dt =.0 0 u 4 = u 4 du =.0 0 dt [separating] =.0 0 t + C [integrating] = u = t + C [multiplying and renaming C] = u = u = (6.0 0 t + C) / [solving] Now we use our initial condition: u(0) = (6.0 0 (0) + C) / = C / = 000, 0

11 which we can solve to get C = 000. Thus, plugging back into our solution, we get u = (6.0 0 t ) /. [Believe it or not, this is equivalent to the answer given in the back of the book.]. #8a Consider an insulated box (a building, perhaps) with internal temperature u(t). According to Newton s law of cooling, u satisfies the differential equation du dt = k[u T (t)], where T (t) is the ambient (external) temperature. Suppose that T (t) varies sinusoidally; for example, assume that T (t) = T 0 + T cos(ωt). Solve the differential equation and express u(t) in terms of t, k, T 0, T, and ω. Observe that part of your solution approaches zero as t becomes large; this is called the transient part. The remainder of the solution is called the stea state; denote it by S(t). We plug in the equation for T (t) and then solve: du dt = k[u (T 0 + T cos(ωt))] [substituting for T (t)] = du dt = ku + kt 0 + kt cos(ωt) [distributing] = du dt + ku = kt 0 + kt cos(ωt) [rearranging] ( ) du = e kt dt + ku = e kt (kt 0 + kt cos(ωt)) [integrating factor with p(t) = k] = (e kt u) = kt 0 e kt + kt e kt cos(ωt) [rewriting each side] = e kt u = T 0 e kt + wekt sin(ωt) + ke kt cos(ωt) ω + k = u = T 0 + kt (w sin(ωt) + k cos(ωt)) ω + k + C [using the integration below] + Ce kt [dividing] Now we justify the integration step above using two steps of integration by parts. In integration by parts, we have two functions u and v and use the formula udv = uv vdu.

12 Warning: this one gets ugly. e kt cos(ωt)dt = e kt cos(ωt)dt [simplifying a bit] [choose u = e kt, dv = cos(ωt)dt = du = ke kt, v = ω sin(ωt)] e kt cos(ωt)dt = ω ekt sin(ωt) k e kt sin(ωt)dt ω [choose u = e kt, dv = sin(ωt) = du = ke kt, v = ω cos(ωt)] e kt cos(ωt)dt = ω ekt sin(ωt) k ( ω w ekt cos(ωt) + k ) e kt cos(ωt)dt ω e kt cos(ωt)dt = ω ekt sin(ωt) + k ω ekt cos(ωt) k ω e kt cos(ωt)dt ) ( + k e kt cos(ωt)dt = ω ekt sin(ωt) + k ω ekt cos(ωt) ω ω + k ω e kt cos(ωt)dt = ω ekt sin(ωt) + k ω ekt cos(ωt) e kt cos(ωt)dt = ωekt sin(ωt) + ke kt cos(ωt) k + ω.4 # For each problem, determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. () (t )y + (ln t)y = t, y() = (5) (4 t )y + ty = t, y() = Let s start with an overview of how to apply Theorem.4.. Here are the steps we can follow: (i) Rearrange the differential equation to the form + p(t)y = g(t). (ii) Find all values of t at which p(t) is not continuous - mark them on a number line with vertical dashes. (iii) Find all values of t at which g(t) is not continuous - mark them on the same number line with vertical dashes. (iv) Mark the t value for which the initial condition is defined as a star on the number line. (v) Theorem.4. is guaranteed to hold on the biggest interval that contains the star and no dashes. In other words, this interval we found is the largest interval in which the solution to the differential equation is certan to exist. Moreover, the solution is unique. Now, let s enumerate the things which cause a function f not to be continuous in this course: (a) f is discontinuous any place where its denominator is equal to 0. (b) f is discontinuous any place where it takes the square root of a negative number. (c) f is discontinuous any place where it takes the ln of a number which is 0 or negative. (d) f is discontinuous any place where it takes sin or cos of a number x with x < or x >. Let s apply this process to our problems.

13 () (i) We rearrange as follows: (t )y + (ln t)y = t = y + ln t t y = t t [dividing] Thus, p(t) = ln t t and g(t) = t t. (ii) We see that p(t) is discontinuous at t <= 0 (because of ln t) and t = (because of the denominator, t ). (iii) We see that g(t) is discontinuous at t = (because of the denominator, t ). (iv) We see that the initial condition (y() = ) is defined at t =. We mark this on the number line with a star. The completed number line appears below. (v) We can see that the biggest interval containing the star and no dashes is (0, ) (which can also be expressed as {t : 0 < t < } - in words, the set of values of t such that 0 < t < ). (5) (i) We rearrange as follows: (4 t )y + ty = t = y + t 4 t y = t 4 t Thus, p(t) = t 4 t and g(t) = t 4 t. (ii) We see that p(t) discontinuous at t = and t = (because of the denominator, 4 t ). (iii) We see that g(t) is discontinuous at t = and t = (because of the denominator, 4 t ). (iv) We see that the initial condition (y() = ) is defined at t =. We mark this on the number line with a star. The completed number line appears below. (v) We can see that the biggest interval containing the star and no dashes is (, ) (which can also be expressed as {t : < t < } - in words, the set of values of t such that < t < )..4 #4,5 In each problem, solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y 0. (4) y = ty, y(0) = y 0 (5) y + y = 0, y(0) = y 0

14 (4) We ll start with solving the differential equation. dt = ty = = tdt [separating] y = y = y = = t + C [integrating] [dividing, renaming C] C t Now we use our initial condition: y(0) = C (0) = C = y 0, which means that C = /y 0. Thus, substituting back in, we get y = /y 0 t. Now note that we can t use Theorem.4. because it is impossible to get the differential equation in the form y + p(t)y = g(t). So, we must simply look at our final answer. We think about where this solution is discontinuous and see that it is discontinuous anywhere that the denominator, /y 0 t, is equal to 0. Solving, we see that this happens when t = ± /y 0. If we think carefully about this, we can realize that this solution exists only when y 0 > 0 (since we can t take the square root of a negative number or divide by 0). So let s look at three cases: (i) y 0 > 0: In this case, y is discontinuous at t = ± /y 0, as we showed above. So the number line appears as follows: Thus, the solution exists on the interval ( /y 0, /y 0 ). (ii) y 0 = 0: In this case, the solution we found above is not even defined, so it s clearly not correct. But we can confirm that the solution y = 0 satisfies the differential equation and the initial condition. Thus, our solution is y = 0, and it exists everywhere (aka on the interval (, )). (iii) y 0 < 0: In this case, the denominator is never equal to 0, so the solution exists everywhere (aka on the interval (, ).). (5) We ll start with solving the differential equation. dt + y = 0 = dt = y [rearranging] = y = dt [separating] = y = t + C [integrating] = y = ± [solving, renaming C] t + C 4

15 Now we use our initial condition: y(0) = ± (0) + C = ± C = y 0. But now we immediately need to examine various cases for y 0 : (i) y 0 > 0: Note that the square root of a number is always nonnegative (if it s even a real number). Thus, if ± C = y 0 and y 0 > 0, then we MUST choose the + sign, not the sign. So, we solve C = y 0 to get C = y0. Plugging back into our solution (and recalling that we chose thet + sign), we get y = t+y 0. Now we must consider where this solution exists. We know that we can only take the square root of a nonnegative number, so we must have t+y 0 know that we cannot divide by 0, so we must have t + y 0. We also 0 0. The first condition requires that t + y0 > 0, which means t > y 0. The second condition can be solve to find t y 0. The number line appears as follows: Thus, the solution exists on the interval ( y0 /, ). (ii) y 0 = 0: In this case, we cannot solve our initial condition equation for C. In fact, our solution for y is impossible since the square of divided by a number can never equal 0. As in problem 4, we can confirm that the solution y = 0 satisfies the differential equation and the initial condition. Thus, our solution is y = 0, and it exists everywhere (aka on the interval (, )). (iii) y 0 < 0: Note that the square root of a number is always nonnegative (if it s even a real number). Thus, if ± C = y 0 and y 0 < 0, then we MUST choose the sign, not the + sign. So, we solve C = y 0 to get C = y0. Plugging back into our solution (and recalling that we chose thet sign), we get y = t+y 0. Now we must consider where this solution exists. We know that we can only take the square root of a nonnegative number, so we must have t+y 0 also know that we cannot divide by 0, so we must have t + y 0. We 0 0. The first condition requires that t + y0 > 0, which means t > y 0. The second condition can be solve to find t y 0. The number line appears as follows: Thus, the solution exists on the interval ( y0 /, ). 5

16 .4 #7 Sometimes it is possible solve a nonlinear equqation by making a change of the dependent variable that converts it into a linear equation. The most important such equation has the form y + p(t)y = q(t)y n, and is called a Bernoulli equation after Jakob Bernoulli. (a) Solve Bernoulli s equation when n = 0; when n =. (b) Show that if n 0,, then the substitution v = y n reduces Bernoulli s equation to a linear equation. This method of solution was found by Leibniz in 696. (a) We ll do both cases of n: (i) [n = 0]: We have the differential equation y + p(t)y = q(t), which is identical to the standard form for the method of integrating factors with g(t) = q(t). Thus, we can use the general solution [Equation ()] given on page 7, namely y = e p(t)dt (ii) [n = ]: We have the differential equation + p(t)y = q(t)y dt = dt = dt [ t ] e p(s)ds g(s)ds + c. t 0 = q(t)y p(t)y [subtracting] = y(q(t) p(t)) [factoring] = = (q(t) p(t))dt [separating] y = ln y = (q(t) p(t))dt + C [integrating] = y = e (q(t) p(t))dt+c [exponentiating] (b) Suppose n 0, and note that if v = y n then v = ( n)y n y, and in particular, y n y = v n. We can find y + p(t)y = q(t)y n = y y n + p(t)y n = q(t) [dividing by y n ] + p(t)v = q(t) [substituting in terms of v] n = v + ( n)p(t)v = ( n)q(t) [multiplying by ( n)] = v We can see that this is a linear equation in terms of v..4 #8 Solve the given Bernoulli equation using the substitution mentioned in Problem 7(b). t y + ty y = 0, t > 0. 6

17 Let s first rearrange this equation to match the form of the equation from problem 7. t y + ty y = 0 = t y + ty = y = y + t y = t y [dividing by t ] Now let s use the transformation from problem 7 (taking n = - so v = y n = y ) and solve: y + t y = t y = v + ( ) t v = ( ) [using the transformation] t = v + 4 t v = t [simplifying] = t 4 ( v + 4 t v ) = t 4 ( t = (t 4 v) = t 6 = t 4 v = 5 t 5 + C [integrating] ) = v = 5t + Ct4 [solving] [integrating factor µ = e ( 4/t)dt = e 4 ln t = e ln(t 4) = t 4 ] Last, we substitute in v = y and solve for y: y = 5t + Ct4 [substituting] = y = [solving] ± 5t + Ct4.5 # This problem involves an equation of the form /dt = f(y). In each problem sketch the graph of f(y) versus y, determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. Draw the phase line, and sketch several graphs of solutions in the ty-plane. First, we sketch a graph of f(y) versus y. /dt = y(y )(y ), y 0 0. Next, we identify the critical/equilibrium points, which are the values of y for which / = 0. These values are y = 0,,. Now we create a phase line. We mark each critical point on the line and then add 7

18 a plus sign for each segment of the graph that is positive, as well as a minus sign for each segment of the graph that is negative. Last, for each plus sign, we draw an arrow to the right, and for each minus sign, we draw an arrow to the left. We identify that y = 0 is unstable, y = is stable, and y = is unstable. Below we sketch a graph of several solutions..5 #, These problems involve equations of the form /dt = f(y). In each problem sketch the graph of f(y) versus y, determine the critical (equilibrium) points, and classify each one asymptotically stable, unstable, or semistable (see Problem 7). Draw the phase line, and sketch several graphs of solutions in the ty-plane () /dt = y (4 y ), < y 0 < () /dt = y ( y), < y 0 < () I will complete the steps as in problem. 8

19 We see that y = is unstable, y = 0 is semistable, and y = is stable. () I will complete the steps as in problem. We see that y = 0 and y = are both semistable. 9

20 .5 #6 Another equation that has been used to model population growth is the Gompertz equation where r and K are positive constants. /dt = ry ln(k/y), (a) Sketch the graph of f(y) versus y, find the critical points, and determine whether each is asymptotically stable or unstable. (b) For 0 y K, determine where the graph of y versus t is concave up and where it is concave down. (c) For each y in 0 < y K, show that /dt as given by the Gompertz equation is never less than /dt as given by the logistic equation..5 # Suppose that a given population can be divided into two parts: those who have a given disease and can infect others, and those who do not have it but are susceptible. Let x be the proportion of susceptible individuals and y the proportion of infectious individuals; then x + y =. Assume that the disease spreads by contact between sick and well members of the population and that the rate of spread /dt is porportional to the number of such contacts. Further, assume that members of both groups move about freely among each other, so the number of contacts is proportional to the product of x and y. Since x = y, we obtain the initial value problem /dt = αy( y), y(0) = y 0, where α is a positive proportionality factor, and y 0 is the initial proportion of infectious individuals. (a) Find the equilibrium points for the differential equation and determine whether each is asymptotically stable, semistable, or unstable. (b) Solve the initial value problem and verify that the conclusions you reached in part (a) are correct. Show taht y(t) as t, which means that ultimately the disease spreads through the entire population..5 # Some diseases (such as typhoid fever) are spread largely by carriers, individuals who can transmit the disease but who exhibit no overt symptoms. Let x and y denote the proportions of susceptibles and carriers, resepectively, in the population. Suppose that carriers are identified and removed from the population at a rate β, so /dt = βy. (5) Suppose also that the disease spreads at a rate proportional to the product of x and y; thus dx/dt = αxy (6) (a) Determine y at any time t by solving Equation (??) subject to the initial condition y(0) = y 0. (b) Use the result of part (a) to find x at any time t by solving Equation (??) subject to the initial condition x(0) = x 0. (c) Find the proportion of the population that excapes the epidemic by finding the limiting value of x as t. 0