Homework 3, due February 4, 2015 Math 307 G, J. Winter 2015

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1 Homework 3, due February 4, 205 Math 307 G, J. Winter 205 In Problems -8, find the general solution of the equation. If initial conditions are given, solve also the initial value problem. is Problem. y + y 2y = 0, y(0) = 2, y (0) =. Solution. Characteristic equation: λ 2 + λ 2 = 0, with roots λ =, 2. The general solution Plug in t = 0: Also, y(t) = C e t + C 2 e 2t y(0) = C + C 2 = 2. y (t) = C e t 2C 2 e 2t y (0) = C 2C 2 =. solving for C and C 2, we get: C = 5/3, C 2 = /3. The solution to the initial value problem: y(t) = 5 3 et + 3 e 2t Problem 2. y 6y + 9y = 0, y(0) = 2, y (0) = 0. Solution. Characteristic equation: λ 2 6λ + 9 = 0, with double root λ = 3. General solution of the equation: y(t) = C te 3t + C 2 e 3t Plug in t = 0: but Solving for C and C 2, we get: Problem 3. 4y 9y = 0. y(0) = C 2 = 2, y (t) = C (e 3t + 3te 3t ) + 3C 2 e 3t y (0) = C + 3C 2 = 0. C = 6, C 2 = 2. y(t) = 6te 3t + 2e 3t Solution. Characteristic equation: 4λ 2 9 = 0, with roots ±3/2. General solution: y(t) = C e 3t/2 + C 2 e 3t/2 Problem 4. 4y + 9y = 0, y(0) =, y (0) = 3.

2 Solution. Characteristic equation: 4λ = 0, with roots ±(3/2)i. General solution: ( ) ( ) 3 3 y(t) = C cos 2 t + C 2 sin 2 t Plug in t = 0: ( ) y(0) = C, y 3 3 (t) = C 2 sin 2 t + 3 ( ) 3 2 C 2 cos 2 t 3 C =, 2 C 2 = 3 C 2 = 2. The solution to the initial value problem is y (0) = 3 2 C 2. cos(3t/2) + 2 sin(3t/2) Problem 5. y + 2y + 3y = 0. Solution. Characteristic equation: λ 2 +2λ+3 = 0, with roots λ = ± 2i. General solution: y(t) = e (C t cos( 2t) + C 2 sin( ) 2t) Problem 6. y + y + y = 0, y(0) = 0, y (0) = 0. Solution. Characteristic equation: λ 2 + λ + = 0, roots General solution: λ = ± 3i. 2 e (C t/2 cos( 3t/2) + C 2 sin( ) 3t/2) Solution to the initial value problem: y(0) = C = 0, y(t) = C 2 e t/2 sin( 3t/2) y (t) = C 2 ( /2)e t/2 sin( 3t/2) + C 2 e t/2 cos( 3t/2)( 3/2), 3 y (0) = 2 C 2 = 0 C 2 = 0. So the solution to the initial value problem is y(t) 0 Problem 7. y + y = 0. Solution. Characteristic equation: λ 2 + λ = 0, with roots λ = 0,. General solution: y(t) = C e 0t + C 2 e t = C + C 2 e t.

3 Problem 8. y = 0. Solution. y = 0 y = C y = C dt = C t + C 2 Problem 9. y + 2y = 0. Solution. λ = 0 λ = ± 2i, and so y = K cos( 2t) + K 2 sin( 2t). Problem 0. Let z = 2 + i, w = 3 4i. Find: z z 2w, w, z, w, arg w, 2 ez. Solution. z 2w = 4 + 9i; z w 2 = 2 + i (3 4i) = 2 + i i + 6i = 2 + i 2 24i 7 = 4 7i + 48i + 24i = z = 2 i w = = 5; arg w = 2π arctan(4/3); e z = e 2 (cos() + i sin()). = (2 + i)( i) ( 7 24i)( i) i. 625 Problem. Two types of molecules, P and Q, interact in a chemical reaction, and make new molecule X: P + Q X. Let p and q, where p < q, be the initial concentrations of P and Q. Let x(t) be concentration of X at time t. Then p x(t) and (t) are concentrations of P and Q at time t, so the rate of reaction is proportional to these concentrations: x (t) = α(p x(t))((t)), where α > 0 is a coefficient. Initially, there was no X, so x(0) = 0. (i) Find x(t). (ii) Find the limit of x(t) as t (without solving the equation). Solution. (i) We have: First, let us find constant solutions: Then divide by (p x)(): dt (p x)() = αdt = α(p x)(). (p x)() = 0 x = p, q. (p x)() = αdt = αt + C.

4 Decompose the fraction inside the integral into partial fractions: We have: So Solving for A and B, we get: Integrate this: So (p x)() = A p x + B. = A() + B(p x) Aq + Bp (A + B)x =. A + B = 0, Aq + Bp =. A = q p, B = q p. (p x)() = ( q p p x ). (p x)() = ( q p ) q p log p x. p x = ( log p x + log ) = q p q p log p x = αt + C. log p x = α(q p)t + C(q p). p x = ec(q p) e α(q p)t. p x = ±ec(q p) e α(q p)t. Let K = ±e C(q p). As C takes any real values, e C(q p) takes any positive real values, and e C(q p) takes any negative real values. So K takes any nonzero real values. We have: p x = Keα(q p)t = Ke α(q p)t (p x) ( Ke α(q p)t ) x = Ke α(q p)t p q. x(t) = Keα(q p)t p q Ke α(q p)t Solve for x(0) = 0. Constants solutions x(t) = p and x(t) = q do not satisfy this condition, so plug in t = 0 into the formula above: 0 = Kp q K K = q p.

5 x(t) = eα(q p)t q q qe α(q p)t /p (ii) This equation has two equilibrium points: x = p and x = q. When x < p, x = α(p x)(q x) > 0, so x increases, and the limit as t is p. (It is useful to draw a diagram.) In Problems 2-3, find general solutions. Problem 2. y = y e t. Solution. This is a linear nonhomogeneous equation. First, solve the corresponding linear homogeneous equation: y = y y = Ce t. Then, use the method of variation of parameters: let C = C(t) and plug in into the original equation: So equating them, we get: y(t) = C(t)e t y = C e t + C(e t ) = C e t + Ce t = Ce t e t. C e t = e t C = C = t + K. The general solution to the nonhomogeneous equation is te t + Ke t Problem 3. t 2 yy = 2. Solution. Separation of variables: t 2 y dy dt = 2 ydy = 2dt t 2 2 y2 = 2 t + C, y = ± 4 t + 2C Problem 4. Newton s law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. If the coffee initially had temperature 00 (in Celcius) in the room with temperature 20, and two minutes later the coffee was 80, find the time (from the beginning) when it cools to 36 (temperature of a human body). Solution. Let y(t) be the temperature at time t. Then Newton s law states that y (t) = α(y(t) 20).

6 This is a linear equation; solving it by usual means, we get y(t) = 20 + Ke αt. Find K: since y(0) = 00, we get K = 80. But y(2) = 80, so e 2α = 80 e 2α = 3 4 α = 2 log(3/4). y(t) = 36 80e αt = 6 e αt = 5 αt = log(/5) t = 2 log(/5)/ log(3/4) =.9 Problem 5. Consider the equation y = y 2 + with the initial condition y(0) = 0. (i) Solve this initial value problem. Find the exact value of y(). (ii) Using Euler s method for h = and h = 0.5, find the approximate value of y(). Solution. (i) Separation of variables: dy y 2 + = dt arctan y = t + C y = tan(t + C). This is the general solution. Find the solution to the initial value problem: 0 = tan C C = 0. So y = tan t is the solution of the initial value problem. y() = tan (ii) For h = : y (0) = y 2 (0) + =, y() y(0) + y (0) = For h = 0.5: y (0) =, y(0.5) y(0) + y (0) 0.5 = 0.5. y (0.5) = + y 2 (0.5) =.25, y() y(0.5) + y (0.5) 0.5 = =.25 Problem 6*. (The Tipping Point.) Suppose that an epidemic starts in a society. Let y(t) be the fraction of infected people at time t. The rate of infection spread is proportional to the quantity of encounters between healthy and infected people, that is, proportional to y(t)( y(t)). That is, for some positive coefficient α, we have: y (t) = αy(t)( y(t)). For the sake of example, suppose α =. Suppose that y(0) = 0.0: initially, there were % infected. (i) Solve the initial value problem. (ii) Find t when y(t) = 30%; when y(t) = 50%; when y(t) = 80%. (iii) Using asymptotic analysis for autonomous equations, find lim t y(t). Solution. (i) Separation of variables: dy y( y) dy = dt y + dy y = dt dy y + dy y = t + C.

7 The left-hand side is equal to log y log y = log y y. log y y = t + C y y = et+c y y = ±ec e t. Now, K = ±e C is an arbitrary nonzero constant. Indeed, as C ranges through all real values, e C ranges through all positive values, and e C ranges through all negative values. Solve for y: y y = Ket y = ( y)ke t y( + Ke t ) = Ke t y = Ket + Ke t. We also forgotten solutions y = 0 and y =, when we divided by y( y). But these solutions are not solutions to the initial value problem y(0) = 0.0. Now, find K by plugging in t = 0: K + K = 0.0 K = ( + K)0.0 K = 99. the solution to the initial value problem is so (ii) y(t) = 99 et + 99 et. y(t) = y 0 Ket + Ke t = y 0 Ke t = ( + Ke t )y 0 K( y 0 )e t = y 0, e t = y 0 K( y 0 ) t = log y 0 K( y 0 ). Plugging in K = /99 and y 0 = 0.3, 0.5, 0.8, we get: t = 3.748; t = 4.595; t = 5.98 (iii) If 0 < y <, then the solution increases, because y = y( y) > 0. And the equilibrium solutions are y = 0 and y =. So the solution starting from y(0) = 0.0 tends to. Problem 7*. Suppose you want to buy a house which costs five times your annual salary. You have already saved money for 20% down payment. You take a 4% 30-year mortgage. The interest is accrued continuously, and your payments are also made continuously. What share of your annual salary should you devote to mortgage payments? Solution. Similarly to the problems from previous hws and lecture 4, let y(t) be the amount of debt you owe at time t. Let us take your annual salary as unit of measure. Let S be the share of salary. Then y (t) = ry(t) S, r = But y(0) = 5 ( 20%) = 4 (initial debt = cost of house - down payment), and y(30) = 0. Solving the equation, we get: y(t) = S r + Cert.

8 We need to find C and S. Using these two initial conditions, we have: Then S = y(0) = S r + C = 25S + C = 4, y(30) = S r + Ce30r = 25S + Ce = 0.