Math 20D Final Exam 8 December has eigenvalues 3, 3, 0 and find the eigenvectors associated with 3. ( 2) det

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1 Math D Final Exam 8 December 9. ( points) Show that the matrix 4 has eigenvalues 3, 3, and find the eigenvectors associated with 3. 4 λ det λ λ λ = (4 λ) det λ ( ) det + det λ = (4 λ)(( λ) 4) + ( λ + ) + ( + λ) = (4 λ)( 3 λ + λ ) + 4λ = 8λ + 4λ + 3λ + λ λ 3 + 4λ = λ 3 + 6λ + 9λ = λ(λ 3). ( ) λ Thus the eigenvalues are 3, 3, and. Perform Guassian elimination on A 3I. Thus the eigenvalues are.. Find the general form of the solution for each of the following differential equations. (a) (5 points) 3y y y = The characteristic polynomial is 3r r = which has roots ± +4 6 = ±5 6. Thus the general form of the solution is c e t + c e 3 t. (b) (5 points) y + 4y + 3y = The characteristic polynomial is r +4r+3 = which has roots 4± 6 5 4±6i = = ± 3i. Thus the general form of the solution is e (c cos(t) + c sin(t)). (c) (5 points) 4y + 4y + y =. Page of 7 Name: Points earned:

2 Math D Final Exam 8 December 9 The characteristic polynomial is 4r +4r+ = which has roots 4± =. Thus the general form of the solution is c e t + c te t. 3. Find the general form of the solution for each of the following systems of differential equations of the form x = Ax. [ (a) (5 points) A = 3 λ det = ( + λ)(3 + λ) + = λ + 4λ λ Thus the eigenvalues are ± i and the eigenvectors are form of the solution is [ [ c e t cos(t) + c cos(t) sin(t) e t. (b) (5 points) A = [ 3 4 [. Thus the general i sin(t) sin(t) + cos(t) λ 3 det = ( λ)( λ) = λ 3λ = (λ 5)(λ + ). 4 λ Thus the eigenvalues are 5 and and the eigenvectors are Thus the general form of the solution is c e 5t 3 + c 4 e t. 3 and, respectively. 4 (c) ( points) A = [ λ 3 det = λ(6 + λ) + 9 = (λ + 3). 3 6 λ Thus the eigenvalue is 3 with multiplicity two and the associated eigenvector is [. Page of 7 Name: Points earned:

3 Math D Final Exam 8 December 9 The generalized eigenvector is a solution to [ 3 3 ξ = 3 3 and so ξ = [ 3. The general solution is then [ [ [ (c t + c )e + c e (5 points) Find the Laplace transform of y satisfying y 5y + 3y = t π π u π(t) + sin(t)u π + e 4t cos() and y() = y () =. The Laplace transform of the left hand side of the equation is (s 5s + 3)L(y) since y() = y () =. Noting that t π = t π π and that sin(t) = sin(t π), the Laplace transform of the right hand side is e πs e πs πs s + e πs s + + s 4. Combining (s 4) +3 these results we get that L(y) = e πs πse πs πs (s 5s + 3) + e πs (s + )(s 5s + 3) + s 4 (s 8s + 5)(s 5s + 3) 5. (5 points) Find y if L(y) = e 3s s +9s 9 s (s +9) + 4 s. First consider the partial fraction decomposition of s +9s 9 s (s +9) Simplifying and equating like terms, we have that A + C = B + D = 9A = 9 9B = 9. = As+B s + Cs+D s +9. Thus A =, B =, C =, and D = 3 and L ( s +9s 9) = t cos() + sin(t). s (s +9) Noting that L ( 4 s ) = 4et we have that y = ( (t 3) cos( 9) + sin( 9)) u 3 (t) + 4e t. 6. ( points) Consider the differential equation y = (y + )(y 4), plot a set of representative solution curves. Note that (y + )(y 4) achieves its minimum at y =. Page 3 of 7 Name: Points earned:

4 Math D Final Exam 8 December 9 Note first that y > for y > 4 and y <, and y < for < y < 4. Further y is concave up when < y < and y > 4 and concave down when y < and < y < 4. Thus we have the following representative solutions with y = being a stable solution and y = 4 being unstable. 6 y(t) t ( points) Write the following differential equation as a system of first order differential equations, y + 3y + y = cos(t) Let x = y, x = y and x 3 = y. Then the differential equation can be defined by x = x x = x 3 x 3 = 3x x + cos(t). 8. (5 points) Plot a set of representative solutions curves for the system of differential [ equations [ x = Ax, where A has eigenvalues and 5 with associated eigenvectors and, respectively. Page 4 of 7 Name: Points earned:

5 Math D Final Exam 8 December 9 4 y -4-4 x ( points) Consider the following situation: There are two tanks of salt water. The first tank starts with 3 gallons and 5 pounds of salt and the second tank starts with gallons of water and no salt. The first tank has a pound per gallon solution flowing into it at a rate of 6 gallons a minute. Two gallons per minute from tank one leave the system while 4 gallons per minute flow from the first tank to the second tank. The second tank has a 3 gallons a minute of fresh water flowing into it and 7 gallons per minute flowing out of the system from tank two. Write a system of differential equations that governs the pounds of salt in each of the tanks. (Note: The volume of liquid in each tank stays constant as a function of time) Let p be the pounds of salt in the first tank and let p be the pounds of salt in the second tank. The rate salt flows into the first tank is 6 lbs gal gal min the rate salt flows out of the first tank is 6 gal min p lbs 3gal. The rate salt flows into the second tank is 3 gal min lbs gal + 4 gal min p lbs gal 3gal. Salt flows out of the second tank at a rate of 7 min p lbs gal. Thus the system is governed by the differential equation p = 6 p 5 p = p 75 7p with the initial conditions p () = 5 and p () =.. (5 points) Give two single variable [ differential equations whose solution can be used to solve e for x and explain. x 4t = Ax + where A = T DT with D = 3e t [ 3 T = [ 3 5 T = [ 5 3 Page 5 of 7 Name: Points earned:

6 Math D Final Exam 8 December 9 Let x = T[ y and then x = T y. Substituting into the differential equation we get e T y = T DT 4t T y + 3e t Then simplifying and multiplying on the left by T we have [ e y = Dy +T 4t. In otherwords, y = 3y +e 4t 5e t and y = y e 4t +9e t. 3e t By solving these two differential equations, we get the vector solution y, but then x = T y and so we have a solution to the original system of differential equations. [ e. Consider the differential equation x 4t = Ax + is c e + c e t. 5e t where the solutions to the homogeneous (a) (5 points) Use variation of parameters to find a particular solution to the above differential equation. [ e e The fundamental matrix is Ψ = t e e t. By the method of variation of parameters we want to find a solution to Ψ e u 4t =. Performing u 5e t Gaussian elimination to find u and u, we get [ e e t e 4t [ e e e e t 5e t t e 4t [ e e 5e t 5e t e 4t t e 4t. 5 e6t Thus u = 5 e6t and u = e 5t + 5 et. And hence u = t 3 e6t + c and u = 5 e 5t + 5 et + c and e t e 4t x = c e t + e t e 4t + t 5 + c e t (b) ( points) Use the method of undetermined coefficients to set up a system of equations that will yield a particular solution to the above differential equation. (Note: Express the system in terms of the matrix A.) Based on the nonhomogenous portion of the differential equation, our particular guess is ae 4t + ( bt + c)e t. Substituting we have that 4 ae 4t + ( b c bt)e t = A ae 4t + A( bt + c)e t + e 4t + e t. 5 Page 6 of 7 Name: Points earned:

7 Math D Final Exam 8 December 9 This yields the following systems of equations 4 a = A a + [ b c = A c + [ 5 b = A b. Page 7 of 7 Name: Points earned: