= 2e t e 2t + ( e 2t )e 3t = 2e t e t = e t. Math 20D Final Review

Transcription

1 Math D Final Review. Solve the differential equation in two ways, first using variation of parameters and then using undetermined coefficients: Corresponding homogenous equation: with characteristic equation y 5y + 6y = e t y 5y + 6y = r 5r + 6 = (r )(3 3) = which has roots r =, r = 3. So we have the fundamental set of solutions y (t) = e t, y (t) = e 3t, and Then our particular solution is y H (t) = c e t + c e 3t. with The Wronskian So that Meaning u = u = With general solution u = e 3s e s e 5s Y(t) = u y + u y y (s)g(s) W(y, y ) ds u = W(e t, e 3t ) = e 5t. ds = e s e s e 5s ds = y (s)g(s) W(y, y ) ds. e s ds = e s u (t) = e t e s = e s u (t) = e t Y(t) = u (t)y (t) + u (t)y (t) = e t e t + ( e t )e 3t = e t e t = e t. y = y H (t) + Y p (t) = c e t + c e 3t + e t.

2 Math D Final Review. Draw the direction field for y = 3 y 3. Solve the IVP: y + 3y = y() = ; y () = 3 Homogenous with constant coefficients. 4. Find the general solution for the system of differential equations: ( ) ( ) x = x + e t 3 The eigenpairs of A are (, ( 3 )) and (, ( )). Corresponding homogenous solution: ( ) ( ) x H = c e t + c e t. 3 There is overlap (e t ) from the homogenous solution to the g(t). Thus the particular solution must have form v = ate t + be t + c Plugging this back into the original DE, we get ( ) ( ) v = te t + β e t + ( ) e t. Note that the β term is contained within x H, so we may disregard it. Thus our final answer is ( ) ( ) ( ) ( ) x = c e t + c e t + te t + e t For the given ODE: (a) Sketch the graph of f (y) vs. y (b) Determine the critical (equilibrium) points (c) Classify the equilibrium points (d) Draw the phase line (e) Sketch several graphs of solutions in the ty-plane dy dt = y(y )(y ), y

3 Math D Final Review 6. Find the Laplace transform of each of the following functions: (a) f (t) = t (b) f (t) = t (c) f (t) = t n for n a positive integer For f (t) piecewise continuous, f (t) Ke at when t M, recall that L{ f (t)} = e st f (t)dt Is the Lapace transform, valid when s > a. Perform the correct checks, take the integral, and we get (a) s (b) s 3 (c) Following the pattern, n! s n+ 7. Find the general solution to the given system of differential equations: 3 4 x = x 4 3 The characteristic equation of the coefficient matrix is 3 r 4 det r = r 3 + 6r + 5r + 8 = 4 3 r Flipping the equation r 3 6r 5r 8 = (r + )(r 7r 8) = (r + )(r + )(r 8) Meaning that we have eigenvalues r = 8 and r = r 3 =. For r = 8 we proceed as usual, finding the eigenvector ξ () =. For r = r 3 =, upon row reducing we find, 3

4 Math D Final Review meaning we have the relationship ξ + ξ + ξ 3 =. Thus we have two free parameters. For this example, let s say that ξ = c and ξ 3 = c 3. Plugging into our equation, we get ξ = (c + c 3 ), and we have ξ = c (c + c 3 ) c 3 = c + c 3 So we get the eigenvectors ξ () = ξ (3) And the general solution x = c e 8t + c e t + c 3 e t 8. (a) Seek power series solutions to the diff eq about the given point x (b) Find the recurrence relation between coefficients (c) Find the first few terms in each of two solutions y, y (d) Evaluate the Wronskian W(y, y )(x ). Check that it is indeed nonzero. y xy y = x = 9. Consider the first order system of equations [ ] x 5 = x 4 (a) Find a fundamental set of solutions to this system, and VERIFY that your constructed functions are fundamental. (b) Sketch the phase portrait of the system, including a few trajectories, and classify the equilibrium at the origin. 4

5 Math D Final Review λ = 6 with ξ () = ( ) and λ = with ξ () = ( 5 ). The fundamental set of solutions via this construction is ( x () e 6t) ( ) ( = = e 6t and x () 5e t ) ( ) 5 = e t = e t e 6t To verify this, we must calculate the Wronskian W(x (), x () ) = det ( x () ( x ()) e 6t 5e = det t ) e 6t e t This is never zero, so we DO have a fundamental set of solutions. Saddle point.. Solve the initial value problem: (6y x + 3) dy dx xy = 3x y() = Method of exact equations. M(x, y) = 3x xy + and N(x, y) = 6y x + 3. Check to see that these are exact. Take integrals: M(x, y)dx = x 3 x y + x + g(y) Equating the pieces, and Ψ(x, y) = y 3 x y + 3y + x 3 + x = c. N(x, y)dy = y 3 x y + 3y + f (x) Plugging in the initial condition, y = when x =, we see that c = 5.. Find the general solution: 9y + 9y 4y = Homogenous, constant coefficients, complex roots.. Solve the system of equations, or show that there is no solution: x + x x 3 = x + x + x 3 = x + x 3 = + x 5

6 Math D Final Review x c x = x = + c = + c x 3 c 3. Consider a population p of pandas that grows at a rate proportional to the current population. That is, dp dt = rp. (a) (a) Find the rate constant r if the population doubles in 3 days. (b) Find r if the population doubles in N days. dp p = rdt ln p = rt + c p(t) = cert Let p be the initial number of pandas present, i.e. p() = p. Then p = p() = ce = c So the population equation is p(t) = p e rt. If the population doubles in 3 days, then p(3) = p() = p, so that p = p e 3r = e 3r r = log 3 (b) r = log N 4. Solve the IVP: y y 3y = 3te t y() =, y () = 5. Transform the given equations into systems of first order equations. (a) u + u + u = (b) u + u + u = 3 sin t (c) u (4) u = 6

7 Math D Final Review (a) Let x = u and x = u. Then x = x x = x x (b) Let x = u and x = u. Then x = x x = 3 sin t x x (c) x = x, x = x 3 x 3 = x 4 and x 4 = x. 6. Find the general solution to the given differential equation. How do solutions behave as t? y + 3y = t + e t Using the method of integrating factors, µ(t) = e p(t), and in this DE, p(x) = 3, so µ(t) = e 3t. So y(t) = µ(s)g(s)ds. µ(t) In our case g(s) = t + e t. y = ce 3t + t e t. As t, the solutions are asymptotic to t 3 9 as t gets large. 7. Find the general solution to the given system of equations: x = x Hint: The characteristic polynomial is (λ + )(λ ) =. 3 λ = yields the eigenvector ξ () = 4. For the repeated eigenvalue, we find only one eigenvector, namely ξ () =. Since we cannot find a second linearly independent vector, we know that our third solution will be of the form x (3) = ξ () te t + ηe t = te t + ηe t. 7

8 Math D Final Review Then (A λi)η = ξ (). Solving this, we find that η = + k. Noting that the k piece is repeated from x () we disregard it and have 3 x = c x () + c x () + c x (3) = c 4 e t + c + c 3 te t + e t 8. Find the general solution: 6y + 4y + 9y = Repeated roots 9. Find the eigenvalues and eigenvectors: ( 3 ) 4 λ = ± i with corresponding eigenvectors ( ) ± i( ). Solve the ODE: y + y sin x = Note that y = would be a solution. In the case where y =, this is a separable system. dy = sin xdx y So y + cos x = c, or y = c cos x. Determine the a n so that the equation n= na n x n + n= a n x n is satisfied. Try to identify the function represented by n= a nx n. 8

9 Math D Final Review Note that when y = n=, this equation is really y + y =. We can solve this using various methods to show that y = ce x. Using series, we first shift our index so that we have [(n + )a n+ + a n ]x n =. n= This forces the recurrence relation a n+ = a n n +. Computing a few terms, we see that in fact Meaning that our solution is y = a n= ( ) n n x n k! a k = ( )k k a k! = a n= ( ) k (x) k k! = a e x.. Determine the longest interval in which this IVP is certain to have a unique solution. Don t try to solve the system. t(t 4)y + 3ty + 4y = ; y(3) = y (3) = This is problem 3..9 from your textbook. Put the equation in standard form, and find any discontinuities of the coefficient functions p(t), q(t), g(t). Then choose the interval where the initial condition lives. This is an application of Theorem 3.. (Existence and Uniqueness Theorem). Recall that standard form is In our case, this becomes y + p(t)y + q(t)y = g(t). y + 4 t 4 y + t(t 4) = t(t 4) There are discontinuities at t = and t = 4. This means that our possible intervals are I = (, ), I = (, 4), and I 3 = (4, ). Note that our initial condition lives in I, so that is our final solution. 3. Determine the order of the given differential equation. Is it linear or nonlinear? If applicable, verify that the given function is a solution to the differential equation. (a) ty y = t ; y = 3t + t (b) d y dt + sin(t + y) = sin t (c) y + 4y + 3y = t; y (t) = t 3 ; y (t) = e t + t 3 (d) d3 y dt 3 + t dy dt + (cos t)y = t 3 9

10 Math D Final Review (a) first order, linear (b) second order, nonlinear (c) fourth order, linear (d) third order, linear 4. Find the general solution to the system of equations [ ] x 3 = x 4 Also sketch a few trajectories, and describe the behavior as t. 5. Consider a tank used in hydrodynamic experiments. After one experiment, the tank contains L of a dye solution with a concentration of g/l. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of L/min, the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches % of its original value. This is.3. from your textbook. To begin with, we decide to let Q(t) measure the amount of dye in the tank, in grams. Note that we started with a concentration of g/l and L of water, so we know Q = Q() = g L = g. L Once we go through the process, we will want to solve for time τ such that Q(τ) = (.) Q() = g. To model the system, we work with the blank rate-in/rate-out model, and fill in the pieces. They key factor in this problem is to notice that the concentration of dye in the exiting water at time t will be the concentration at that, which is the amount at that time divided by the total about of water in the tank. That is, Q(t). Q t g min = rate in g g rate out min ( min concentration in = (flow rate in) entering water ) (flow rate out) dq dt = L min gl L min Q(t)g L = Q(t) g min ( ) concentration in exiting water

11 Math D Final Review Now we ve come up with a differential equation: dq dt = Q. We solve this, and recover the general solution Q(t) = ce t Plugging in our initial condition Q() =, we see such that our specific solution is = Q() = ce = c Q(t) = e t Now we can attack the real question: for what time τ is Q(τ) =? Which is our final solution. Q(τ) = = e τ. = e τ ln(.) = τ τ = ln(.) 6. Determine the general solution of the system 5 x = 5 x. 5 Hint: The characteristic polynomial of this matrix is (λ 3)(λ 6). You may use this fact without proving it, i.e. you do not need to take the determinant of a 3x3 matrix. This is exactly like the exam problem. The eigenvalues are different, but the vectors are the same. 7. Solve the IVP dy dt + y3 = ; y() = y What values of y are acceptable? How does our choice of y dictate the interval of t on which the DE is defined? This is.4.4 in your textbook. Note that this is non-linear. It is, however, separable. We get y(t) = y y t +