Review for Exam #3 MATH 3 Lodwick/Kawai You will have hrs. to complete Exam #3. There will be one full problem from Laplace Transform with the unit step function. There will be linear algebra, but hopefull, it will not require a ton of arithmetic. Otherwise, everthing is from Chapters 5 & 9. (#) Solve this LT problem: Solve:, t < g (t) t, t + g (t), (), (). (#) Solve b elimination: (D 3) u + (D ) v t (D + ) u + (D + 4) v (#3) Be sure that ou can set up a simple sstem like Section 3., #3. Answer is in back of the book. (#4) Find all critical points for this sstem: (#5) The origin is the onl critical point. dx/dt x + d/dt x Determine the nature of the trajectories of the various solutions: Hint: This should be exact. dx/dt x + d/dt (#6) Solve the logistic separable: What happens as t? d dt (4 ), (). (#7) Suppose we did have a 3 3 sstem A with: A (a) What are the 3 eigenvalues?
(b) For the repeated eigenvalue, are there TWO families of eigenvectors when we solve: (A λi) v (c) If so, then write the general solution with c, c, and c 3 in it. (t) (t) c??? + c??? + c 3??? 3 (t) (#8) Find the eigenvalues and eigenvectors for: A 5 3 (a) Graph them on the phase plane. (b) Do all the solutions converge to: (t) (t)? (#9) Solve the IVP for A: A 3, () () (#) Give the form of particular solution p. Do not solve. A λ,,. and the sstem is: A t t cos (t) 3 (#) Evaluate the Wronskian if our three independent solutions to a 3 3 homogeneous problem are: e t e t e t e t e t, e t, 3 e t Are the three solutions independent? What were the eigenvalues for this sstem and what are their corresponding eigenvectors? If we had told ou that these would form a fundamental matrix that is the solution to: (t) (t) A (t) (t) 3 (t) 3 (t)
So, if we tell ou that the fundamental matrix is: e t e t e t X e t e t and X e t e t 3 e t e t e t e t e t e t e t e t e t, then can ou plug into this formula? Particular solution is p X X e t dt Show how we might set up the particular solution to this (assuming that we know the eigenvalues and eigenvectors above): e t + Short Solutions (#) Here is a graph of g (t). 3 t At t, we need to turn off t, and turn on ( t). Thus, g (t) t + ( t + t) u (t ) g (t) t + ( t) u (t ) g (t) t (t ) u (t ) We can use one of the special LT transforms! L { } s Y (s) s () () s Y (s) L {} Y (s) Thus, the left side is: ( s + ) Y (s) The right side is: L {t} L {(t ) u (t )}! s ( e s L {t} ) s e s The original function was f (t) t and f (t ) t. We could also use the right-shift transform: L {g (t) u (t )} e s L {g (t + )} ( ) s. In this case, we have g (t) t and g (t + ) t, and we get the same result. 3
Our LT equation is: ( s + ) Y (s) s e s ( s ) ( s + ) Y (s) + s e s ( s ) It turns out that we are better off combining the first two fractions on the right side: ( s + ) ( ) Y (s) s + s e s s because now we can divide through b ( s + ). Y (s) ( ) s e s s (s + ) Now comes the unfortunate linear algebra: Multipl through b s ( s + ). s (s + ) A s + B s + Cs + D s + As ( s + ) + B ( s + ) + Cs 3 + Ds There is onl one magic number: Let s. B. Now we must balance the polnomial. As ( s + ) + ( s + ) + Cs 3 + Ds s 3 + s + s + As 3 + As + ( s + ) + Cs 3 + Ds There is onl one s term, and the left side coeffi cient is zero, so A. s 3 + s s + Cs 3 + Ds Excellent. We see that C must equal zero, and then D. Our partial fraction decomposition is: s + s + s s + s s + Thus, we have: Y (s) ( s e s s ) s + We use the identit in reverse. This is eas. L { e s ( s )} (t ) u (t ) f (t) t. 4
{ ( )} L e s s sin (t ) u (t ) f (t) sin (t). + Our final answer (with coeffi cients) is: (t) t (t sin (t )) u (t ). This is a little too much work for an exam, but I will give ou one with fewer steps. The principle will still be the same. 3 4 5 x Since g (t) was continuous, we expect the solution curve to be continuous AND it should be smooth at t. This is a spline solution! 4 (#) To be announced soon. (#3) Be sure that ou can set up a simple sstem like Section 5., #3. Answer is in back of the book. ( ) ( x ) x. + 7 Organize: ( x ) ( ) 3 3 x.7x +. +..3x.3 If we want the long-term equilibrium solution, we set both equal to zero:.7x +. +..3x.3 Not surprisingl, the equilibrium solutioni is x and. The would make the concentration in both tanks equal to. kg/l, which is the concentration of the input pipe into Tank A. (#4) Find all critical points for this sstem: dx/dt x + d/dt x 5
This give us the circle: and the two intersecting lines: x + x ±x. Thus, we have four interesections: ( ) (,,, ), (, ), and (, ). (#5) The origin is the onl critical point. Determine the nature of the trajectories of the various solutions: Hint: This should be exact. d dx ( ) x + dx/dt x + d/dt dx (x + ) d dx + (x + ) d We have: Mdx + Nd is exact iff N x M. Integrate both terms: F dx x (x + ) d x + So the famil of solutions is: The look like this: x + C. x + 4 The are hperbolas. The solid line is C and the dashed line is C. 4 4 x 4 6
(#6) Solve the logistic separable: Again partial fractions: This gives us: ( ln 4 ( 4 ) d dt (4 ) A + + 4 (4 ), (). d (4 ) dt B 4 ( ) 4 + ( ) 4 4 ) d dt 4 (ln () ln (4 )) t + C 4t + C 4 C 3e 4t As t, we should reach an equilibrium value of 4. e 4t + C 3 e 4t e 4t + C 3 4 C 3 C 3 4 e 4t + (/4) 4 + 4e 4t. (#7) Suppose we did have a 3 3 sstem A with: A (a) What are the 3 eigenvalues? It s diagonal, so we have λ,,. (b) For the repeated eigenvalue, are there TWO families of eigenvectors when we solve: (A λi) v Solve for λ : ( ) ( ) ( ) We can onl be sure that x 3. We have x and x are free variables, so we tpicall assign: v and v x x x 3 x x x 3 In theor, we could assign an independent linear combination of these vectors, but this makes it eas. 7
(c) If so, then write the general solution with c, c, and c 3 in it. (t) (t) 3 (t) c e t + c e t + c 3 e t c e t c e t c 3 e t (#8) Find the eigenvalues and eigenvectors for: 5 A 3 5 λ 3 λ ( 5 λ) ( λ) + 6 λ + 5λ + 6 (λ + 3) (λ + ) λ 3,. For λ 3, we solve: 5 ( 3) 3 ( 3) 3 3 x We have: x + x, so we choose v For λ, we solve: 5 ( ) 3 ( ) 3 3 x We have: 3x + 3 x Clearl, these vectors will have slope m 3/, and we choose x. v 3/ (a) Graph them on the phase plane. Here are the two lines: 4 4 4 4 x 4 4 x 4 4 The associated vector field looks like the diagram on the right. All trajectories lead to (, ). 8
(b) Do all the solutions converge to: (t) (t)? YES. The eigenvalues are negative, so e t and e 3t. (#9) Solve the IVP for A: A 3, () () We need a fundamental matrix. Eigenvalues: λ 3 λ (λ ) (λ + ) 3 λ 4 λ,. For λ, we solve: ( ) 3 ( ) For λ, we solve: 3 x x 3 3 3 3 The slopes of the vectors are ( ) and (/3). x x v v 3 The fundamental matrix is: and the IVP solution is the solution to: e t e t e t e t 3e t e t 3e t e t c c The inverse of X is: and the solution is (at t ): c 4 c ( 3) 3 e t e t 4 3e t e t / /. We have: e t e t 3e t e t / /. 9
(#) Give the form of particular solution p. Do not solve. A λ,,. and the sstem is: A t t cos (t) 3 Man undetermined coeffi cients. We see that the left-side roots are λ,, and these do not conflict with an of the non-homogeneous parts: λ,, ±i, ±i. We need to solve for SIX VECTORS: p a + tb + (c + td) cos (t) + (e + tf) sin (t). (#) We have: e t e t e t p e t e t e t e t 3 3 e 3t te t (/3) e t te t (/3) e t te t (/3) e t dt 3 3 te t e t e t e t e t e t e t e t 9 e t ( /3) e 3t t t Clearl, the terms with te t are new and independent of our fundamental solutions. The vector T is independent of the other two vectors associated with e t :, We can test this with the Wronskian and determinant is non-zero. 3. Thus, if we have the abilit to invert a 3 3 matrix easil, then this must be less linear algebra work than the Method of Undetermined Coeffi cients. We didn t get to Variation of Parameters in Sect. 9.7, but we are essentiall using the same principle as in Sect. 4.6! The Undetermined Coeffi cients set-up would be: p ae t + bte t because λ conflicts with roots from the left side matrix. solve. This would be diffi cult to