20D - Homework Assignment 4

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1 Brian Bowers (TA for Hui Sun) MATH 0D Homework Assignment November, 03 0D - Homework Assignment First, I will give a brief overview of how to use variation of parameters. () Ensure that the differential equation takes the form y + p(t)y + q(t)y g(t) where p, q, and g are continuous. [If we don t have this form and we can t get to this form, then we CAN T USE THIS METHOD.] () Find a fundamental set of solutions y and y to the associated homogeneous differential equation (namely y + p(t)y + q(t)y 0). (3) Calculate W (y, y ) y y y y. () Calculate Y y (5) The general solution is y c y + c y + Y. y g W (y, y ) dt + y y g W (y, y ) dt. 3.6 #7,0 For each problem, find the general solution of the given differential equation. (7) y + y + y t e t, t > 0 (0) y y + y e t /( + t ) (7) y + y + y t e t, t > 0. I will use the method outlined above: () We have the right form, and we see that p, q, and g are continuous [since t > 0]. () The homogeneous equation is y + y + y 0. This has characteristic equation r + r + 0, which factors as (r + ) 0, yielding solutions r,. Thus, y e t and y te t. (3) W (y, y ) y y y y e t (e t te t ) ( e t )te t e t te t + te t e t. () We calculate: Y y y g W (y, y ) dt + y te e t t (t e t ) e t dt + te t e t t dt + te t t dt e t ln t + te t ( t ) e t ln t e t [since t > 0] y g W (y, y ) dt e t (t e t ) dt e t (5) The general solution is y c y + c y + Y c e t + c te t e t ln t e t c e t + c te t e t ln t [combining the e t terms]

2 (0) y y + y e t /( + t ). I will once again use the method outlined above: () We have the right form, and we see that p, q, and g are continuous [since + t is never equal to 0]. () The homogeneous equation is y y + y 0. This has characteristic equation r r + 0, which factors as (r ) 0, yielding solutions r,. Thus, y e t and y te t. (3) W (y, y ) y y y y e t (e t + te t ) e t (te t ) e t + te t te t e t. () We calculate: y g Y y W (y, y ) dt + y y g W (y, y ) dt te e t t (e t /( + t )) e e t dt + te t t (e t /( + t )) e t dt e t t dt + tet + t + t dt et ln( + t ) + te t tan t (5) Thus, the general solution is y c y + c y + Y c e t + c te t et ln( + t ) + te t tan (t) 3.6 #3,7 In each problem, verify that the given functions y and y satisfy the corresponding homogeneous equation; then find a particular particular solution of the given nonhomogeneous equation. (3) t y y 3t, t > 0; y (t) t, y (t) t (7) x y 3xy + y x ln x, x > 0; y (x ), y (x)x ln x (3) First, we confirm that y and y satisfy the differential equation by plugging them into the left side of the differential equation and confirming that it equals 0: t y y t (t ) (t ) t (t) (t ) t () t 0 t y y t (t ) (t ) t ( t ) t t (t 3 ) t 0 Next, we follow steps through from above: (a) We don t have the right form yet, so we divide both sides by t to get y t y 3 t Now we see that p, q, and g are continuous [since the problem statement tells us t > 0]. (b) This step was completed for us in the problem statement. (c) W (y, y ) y y y y t ( t ) (t)t 3. [Note that since W 0, we know that y and y form a fundamental set of solutions.]

3 (d) We calculate Y y y g W (y, y ) dt + y y g W (y, y ) dt t t (3 t ) t dt + t (3 t ) dt 3 3 t (t 3 /3 t )dt + t (/3 t )dt t t t ln t + t 6 3 t t ln t + 3 t 3 + t ln t t 3 We note that Y (t) is a particular solution. However, in the solution, they have combined the t 3 with y to yield the particular solution term Y (t) + t ln t (7) x y 3xy + y x ln x, x > 0; y (x) x, y (x) x ln x. First, we confirm that y and y satisfy the differential equation by plugging them into the left side of the differential equation and confirming that it equals 0: x y 3xy + y x (x ) 3x(x ) + (x ) x (x) 3x(x) + x x () 6x + x 0 x y 3xy + y x (x ln x) 3x(x ln x) + (x ln x) x (x ln x + x) 3x(x ln x + x) + x ln x x ( ln x + + ) 6x ln x 3x + x ln x x ln x + x + x 6x ln x 3x + x ln x 0 Next, we follow steps through from above: (a) We don t have the right form yet, so we divide both sides by x to get y 3 x y + x y ln x Now we see that p, q, and g are continuous [since the problem statement tells us x > 0]. (b) This step was completed for us in the problem statement. (c) W (y, y ) y y y y x (x ln x + x) (x)x ln x x 3 ln x + x 3 x 3 ln x x 3. [Note that since W 0, we know that y and y form a fundamental set of solutions.] 3

4 (d) We calculate y g Y y W (y, y ) dx + y y g W (y, y ) dx x x ln x(ln x) x x 3 dx + x (ln x) ln x x 3 dx (ln x) x ln x dx + x ln x x x dx [ ] [ ] x 3 (ln x)3 + x ln x (ln x) 3 x (ln x) 3 + x (ln x) 3 6 x (ln x) 3 7. # Transform the given equation into a system of first order equations: u + 0.5u + u 0. In general, to convert a second order differential equation into a system of first order differential equations, we choose to set x equal to the dependent variable (u in this case) and then we set x x. (In this case, it means that x x u. Furthermore, we see that u (u ) (x ).) So if we plug u x, u x, and u x into our original differential equation, we get x + 0.5x + x 0. Thus, our system of first order equations is { x + 0.5x x 0 x x. 7. #5 Transform the given initial value problem into an initial value problem for two first order equations. u + 0.5u + u cos(3t), u(0), u (0). We use the same substitutions as in the previous exercise (u x, u x, and u x ) to get the system { x + 0.5x + x cos(3t) x x, x (0), x (0). 7. #7 Systems of first order equations can sometimes be transformed into a single equation of higher order. Consider the system x x + x, x x x. (a) Solve the first equation for x and substitute into the second equation, thereby obtaining a second order equation for x. Solve this equation for x and then determine x also. (b) Find the solution of the given system that also satisfies the initial conditions x (0), x (0) 3. (c) Sketch the curve, for t 0, given parametrically by the expressions for x and x obtained in part (b). (a) Solving the first equation for x, we get x x + x.

5 Substituting this into the second equation ad solving for x, we get (x + x ) x (x + x ) x + x x x x x + x + 3x 0 r + r [characteristic equation] (r + 3)(r + ) 0 [factoring] r 3, x c e 3t + c e t Next, we plug this into our equation for x to get x (c e 3t + c e t ) + (c e 3t + c e t ) 3c e 3t c e t + c e 3t + c e t c e t c e 3t (b) Now we use the initial conditions: x (0) c e 3(0) + c e (0) c + c x (0) c e (0) c e 3(0) c c 3 The first equation yields c c. Plugging this into the second equation, we get c ( c ) c 3, which we can solve to get c 5/. Plugging this back into the first equation, we get c + 5/, which we can solve to get c /. Thus, we plug these constants into our formulae for x and x : x e 3t + 5 e t x 5 e t + e 3t (c) We can get some intuition about the graph by thinking about what happens at t 0 and what happens for very large values of t. At t 0, we have x x (0) and x x (0) 3, so we graph the point (, 3). At large values of t, we see that both x and x will approach 0. By experimentally plugging in several values of t, we can see that the graph looks like the following: 5

6 7. #, For each problem, proceed as in problem 7. (a) Transform the given system into a single equation of second order. (b) Find x and x that also satisfy the given initial conditions. (c) Sketch the graph of the solution in the x x -plane for t 0. () x x, x (0) 3, x x, x (0). () x 0.5x + x, x (0), x x 0.5x, x (0). () We complete the steps in order: (a) First, we solve the first equation for x to get x x. Next, we plug this into our second equation to get x x x x x + x 0 (b) Next, we solve this equation for x : x + x 0 r + 0 [characteristic equation] r ±i [solving using quadratic formula] x c cos(t) + c sin(t) Plugging this back into our equation for x, we see x x (c cos(t) + c sin(t)) c sin(t) + c cos(t). Now we use our initial conditions: x (0) c cos((0)) + c sin((0)) c 3 x (0) c sin((0)) + c cos((0)) c Finally, we plug c and c back into our formulae for x and x : x 3 cos(t) + sin(t) x 3 sin(t) + cos(t) (c) We can find by experimentation that this actually graphs a circle (traversed clockwise) centered at 0 of radius 5. (We can confirm this by calculating (x, x ) 5.) 6

7 () x 0.5x + x, x (0), x x 0.5x, x (0) We complete the steps in order: (a) First, we solve the first equation for x to get x x + x. Next, we plug this into our second equation to get ( x + ) ( x x 0.5 x + ) x x + x x x 8 x x + x x 0 (b) Next, we solve this equation for x : x + x x 0 r + r [characteristic equation] r ± i x c e t/ cos(t) + c e t/ sin(t) Plugging this back into our equation for x, we see x x + x ( ) c e t/ cos(t) + c e t/ ( ) sin(t) + c e t/ cos(t) + c e t/ sin(t) ( c e t/ cos(t) c e t/ sin(t) c ) e t/ sin(t) + c e t/ cos(t) + ( ) c e t/ cos(t) + c e t/ sin(t) ( c c + c + c ) ( e t/ cos(t) + c c + c ) e t/ sin(t) c e t/ cos(t) c e t/ sin(t) Now we use our initial conditions: x (0) c e (0)/ cos((0)) + c e (0)/ sin((0)) c x (0) c e (0)/ cos((0)) c e (0)/ sin((0)) c 7

8 Finally, we plug c and c back into our formulae for x and x : x e t/ cos(t) + e t/ sin(t) x e t/ cos(t) + e t/ sin(t) (c) We can note that x and x are decreasing over time, so our graph should approach the origin. In particular, if we experiment with plugging in various values of t, we see that the graph will spiral inward clockwise, as seen below: 7. #3 Transform the following equations for the parallel circuit into a single second order equation: di dt V L, dv dt I C V RC, where L is the inductance, C is the capacitance, and R is the resistance. Note that I (the current) and V (the voltage) are the dependent variables here, whereas the other letters are constants. We begin by solving the first equation for V : di dt V L Next, we plug this into our second equation: V L di dt. dv dt I C V RC d ( L di ) IC L di dt dt dt RC L d I dt I C L di RC dt LRC d I dt LRC d I dt + LdI dt + RI 0 LRCI + LI + RI 0 IR LdI dt 7. # If A (a) A B (b) 3A + B (c) AB (d) BA + i + i i 3 and B, find 3 + i i i 8

9 (a) + i + i i 3 A B 3 + i i i + i + i i i i i + i i + i i i + i i 7 + i + i + 3i (b) + i + i i 3 3A + B i i i 3 + 3i 3 + 6i i i 6 3i i 3 + 3i + i 3 + 6i i + 6 3i i 3 + i 6i + 6i 6 5i (c) + i + i i 3 AB 3 + i i i ( + i)(i) + ( + i)() ( + i)(3) + ( + i)( i) (3 + i)(i) + ( i)() (3 + i)(3) + ( i)( i) i + i 3 + 3i + i + 3i + i 9 + 6i i 3 + 5i 7 + 5i + i 7 + i (d) i 3 + i + i BA i 3 + i i (i)( + i) + (3)(3 + i) (i)( + i) + (3)( i) ()( + i) + ( i)(3 + i) ()( + i) + ( i)( i) i i i + 6 3i + i 6i + + i i 8 + 7i i 6 i 7. #0 Either compute the inverse of the matrix or show that it is singular: ( ) 3 9

10 Call the above matrix A. First, we quickly check whether or not the matrix is singular. We see that det(a) (3) ( ) 0, so the matrix is invertible. Now we augment the matrix with the identity and use row operations to convert the left side to the identity matrix. 0 [augmenting with identity matrix] [adding *(row ) to (row )] 0 0 [dividing (row ) by ] [adding -*(row ) to (row )] Thus, the inverse is the right side of our augmented matrix, namely ) ( 3 7. #,3 In each problem, verify that the given vector satisfies the given differential equation. ( 3 () x x, x e ) t. ( ( ( (3) x x + e 3 ) t, x e 0) t + te ) t. e t () First, we simplify the expression for x as x e t. Now we substitute in and confirm that the equation is true: e t 3 e t e t e t 8e t 3(e e t t ) (e t ) (e t ) (e t ) 8e t 8e t e t e t Thus, the given x satisfies the differential equation. ( e (3) First, we simplify the expression for x as x t + te t te t equation is true: e t + te t e te t t + te t e t 3 te t + e t e t + e t + te t (e e t + te t t + te t ) (te t ) 3(e t + te t ) (te t + 3e t + te t e e t + te t t + te t e t 3e t + te t + e t 3e t + te t 3e e t + te t t + te t e t + te t ). Now we substitute in and confirm that the ( e t e t ) 0

11 Thus, the given x satisfies the differential equation.