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Day 4 Differential Equations (option chapter) Recall from AP Calculus The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present (at least for awhile.) So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account. If the rate of change is proportional to the amount present, the change can be modeled by: dy ky dt dy ky dt 1 dy =k y dt 1 y dy k dt Rate of change is proportional to the amount present. Divide both sides by y. Integrate both sides. ln y = k t +C 1 y dy k dt Integrate both sides. ln y = k t +C e ln y =e kt + C y = e ×e C kt Exponentiate both sides. When multiplying like bases, add exponents. So added exponents can be written as multiplication. ln y e Exponentiate both sides. kt C e y e e C When multiplying like bases, add exponents. So added exponents can be written as multiplication. kt y e e C kt y Ae kt Since eC is a constant, let e C A. y e e C kt y Ae kt y0 Ae Since eC is a constant, let e C A. 1 k 0 At t 0 , y y0 . y0 A y y0 e kt This is the solution to our original initial value problem. So if we start with: We end with: dy ky dt y y0 e kt What if we have a series of differential equations? dy1 = ky1 dt dy2 = ky2 dt dy3 = ky3 dt We could solve these individually y1 =c1ekt y2 =c2ekt y3 =c3ekt Provided that we have initial conditions for each of these to solve for the constants. If we define x x’(t) = [ ] … x1’(t) x2’(t) xn ’(t) This yields the equation x’(t)= Ax Which is easy to solve in the case of a diagonal matrix. [ ] [ ][ ] x’1 x’2 x’3 = 3 0 0 0 -2 0 0 0 4 x1 x2 x3 We can solve each of these as a separate differential equation x1’ = 3x1, x2’ = -2x2, x3’ = 4x3 x1 (t) = b1e3t, x2 (t) = b2e-2t, x3 (t) = b3e4t, This is the general solution. We can solve for the constants if given an initial condition. First order homogeneous linear system of differential equations x1’(t) = a11x1 (t) + a12 x2 (t) + … a1nxn (t) x2’(t) = a21x1 (t) + a22 x2 (t) + … a2nxn (t) … xn ’(t) = an1x1 (t) + an2 x2 (t) + … annxn (t) We could write this in matrix form as: x1’ (t) a11 a12 …. a12 x(t) = x2’ (t) A = a21 a 22 … a2n xn’ (t) … … [] [ ] an1 a n2 …anm What if our system is not diagonal? The system at the left can be written as du/dt = Au with a as du1 = -u1 + 2u2 dt [ ] A = -1 2 1 -2 du2 = u1 – 2u2 dt [] Initial condition u(0) = 1 0 How can we solve this system? du/dt = Au y = eAt u(t) = c1eλ1 t x1 +c2e λ2 t x2+…+ cneλ n t xn Check that each piece solves the given system du/dt = Au d (eλ t x1) = A eλ t x1 dt λeλ t x1 = A eλ t x1 λx1 = Ax1 Key Formulas Difference Equations Differential Equations du/dt = Au y = eAt Solve the differential equations The system at the left can be written as du/dt = Au with a as [ ] A = -1 2 1 -2 Start by computing the eigenvalues and eigenvectors What are the eigenvalues from inspection? Hint: A is singular The trace is -3 Solve the differential equations Step 1 find the eigenvalues and eigenvectors We can a solve via finding the determinant of A - λI By inspection: the matrix is singular det -1-λ 2 therefore 0 is an eigenvalue the trace is 1 -2-λ -3 therefore the other eigenvalue is -3. [ ] Calculate the eigenvector associated with λ = 0, -3 [ ] [ ] A = -1 2 1 -2 A+ 3I = 2 1 2 1 For λ = 0 find a basis for the kernel of A [] [] 2 1 For λ= -3 find a basis for the kernel of A+3I 1 -1 Solve the differential equations The system at the left can be written as du/dt = Au with a as Note: the solutions of the equations are going to be e raised to a power. [ ] A = -1 2 1 -2 The form that we are expecting for the answer is y = c1 e λ t x1 + c2 e λ t x2 1 2 The eigenvalues are already telling us about the form of the solutions. A negative eigenvalue will mean that that portion goes to zero as x goes to infinity. An eigenvalue of zero will mean that we will have an e0 which will be a constant. We will call this type of system a steady state. Solve the differential equations Solve by plugging in eigenvalues into expected equation and for λ1 and λ2. and the corresponding eigenvectors in x1 and x2 [ ] A = -1 2 1 -2 y = c1 e0t 2 + c2 e -3t 1 1 -1 [] We find c1 and c2 by using the initial condition Recall: Initial condition u(0) = 1 0 c1 = 1/3 c2 = 1/3 [] Plugging in zero for t and the initial conditions yields: [] [] [ ] [ ] 1 0 = c1 2 1 + c2 1 -1 Solve the differential equations The general solution is y = 1/3 2 + 1/3 e -3t 1 [] [] 1 -1 We are interested in hat happens as time goes to infinity Recall our initial condition was 1 all of our quantity was in u1 0 [] Then as time progressed there was flow from u1 to u2. As time approaches infinity we end with the steady state 2/3 1/3 [] The solution to y’ = ky is y = y0ekt The solution to x’ = Au is u = c0eAt Homework: wkst 8.4 1-9 odd, 2 and 8 What if the matrix is not diagonal? • White book p. 520 ex 3, 4, 5