Math 308 Week 8 Solutions

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1 Math 38 Week 8 Solutions There is a solution manual to Chapter 4 online: math/. This online solutions manual contains solutions to some of the suggested problems. Here are solutions to suggested problems that cannot be found in the online solutions manual. Matlab 5 In exercises -6 perform the following tasks for the given initial value problem: a) Find the exact solution. b) Use MATLAB to plot on a single figure window the graph of the exact solution, together with the plots of the solutions using each of the three methods (,, ) with the given step size h. Use a distinctive marker (type help plot) for help on available markers) for each method. Label the graph appropriately and add a legend to the plot.. x = x sin(3t), with x() =, on [, 4]; h =. Answer: First, we use dsolve to find the solution. We get: x(t) = e cos(3t)/3 cosh(/3) sinh(/3) Next, we have Matlab plot the above solution: >> t = :.:4; >> g = inline('exp(-cos(3*t)/3)/(cosh(/3)-sinh(/3)','t'); >> plot(t, g(t), '-'); The third entry in the plot command says to use a solid line for this graph. Now, we use,, and. Here are the Matlab commands: >> f= inline('x*sin(3*t)', 't', 'x'); >> [t, x] = (f, [, 4],,.); >> [t, x] = (f, [, 4],,.); >> [t3, x3] = (f, [, 4],,.); Now, we plot the results on the same figure as the plot of the solution. We will use dotted lines for these graphs, and we will include a legend: >> hold on >> plot(t, x, ':'); >> plot(t, x, '-.'); >> plot(t3, x3, '--'); >> legend('', '', '', '', 3); >> shg

2 The final entry in the legend command says which corner to put the legend in. Here is the resulting graph: If instead of using dotted lines, we wish to have colored lines, we can enter the commands: >> hold on >> plot(t, x, 'r'); >> plot(t, x, 'g'); >> plot(t3, x3, 'k'); >> legend('', '', '', '', ); >> shg

3 . y = ( + y ) cos(t), with y() =, on [, 6]; h =.5 Answer: We use the same method as in problem. The solution to the initial value problem is: The resulting graphs are: y = tan(sin t)

4 3. z = z cos(t), with z() =, on [, 6]; h =.5 Answer: We use the same method as in problem. The solution to the initial value problem is: The resulting graphs are: y = cos t sin t

5 4. x = x(4 x)/5, with x() =, on [, 6]; h =.5 Answer: We use the same method as in problem. The solution to the initial value problem is: The resulting graphs are: x = 4 + 3e 4t/

6 5. y = ty, with y() =, on [, ]; h =. Answer: We use the same method as in problem. The solution to the initial value problem is: The resulting graphs are: y = e t

7 6. z = ( z)t, with z() = 3, on [, ]; h =.5 Answer: We use the same method as in problem. The solution to the initial value problem is: The resulting graphs are: z = + e t3 /

8 . Simple substitution will reveal that x(t) = π/ is a solution of x = e t cos x. Moreover, both f(t, x) = e t cos x and f/ x = e t are continuous everywhere. Therefore, solutions are unique and no two solutions can share a common point (solutions cannot intersect). Use Euler s method with a step size h =. to plot the solution of x = e t cos x with initial conditions x() = on the interval [, π] and hold the graph (hold on). Overlay the graph of x = π/ with the command line([, *pi], [pi/, pi/], 'color', 'r') and note that the solutions interesect. Does reducing the step size help? If so, does this reduced step size hold up if you increase the interval to [, ]. How do and perform on this same problem? Answer: First, we perform Euler s method with step size h =.. The commands are: >> f = inline('exp(t)*cos(x)', 't', 'x') >> [t, x] = (f, [,*pi],,.); >> plot(t, x) We see that it gets fairly messed up towards the end. We can plot the line x = π/ on the same figure: >> hold on >> line([, *pi], [pi/, pi/], 'color', 'r') We get the following graph:

9 We see that the Euler s method solution is intersecting the line x = π/. If you try reducing the step size, you will eventually find a step size for which the result of Euler s method does not intersect the line x = π/. For example, the step size h =.5 works, and gives the following graph: However, this step size does not work on the interval [, ]:

10 If you try and, you will see that they both have the same problem. Here is a graph of on the interval [, π] with step size h =.: Here is a graph of on the interval [, ] with step size h =.:

11 For all of the methods (,, and ), you can choose your step size small enough to work on any given interval. However, if you expand the interval, you will need a smaller step size, and some intervals may require such a small step size that the computation time becomes unreasonable. If you look at the direction field of this differential equation in dfield7, you will see why this is occurring. The slopes of solutions to the differential equation are incredibly steep near for most points with t > 3 (of course, the slopes approach at x = π/, but you have to be really, really close to x = π/ to get to the slopes that are approaching ). In dfield7, you can ask Matlab to compute the solution with initial condition x() = on the interval from t = to t =. It successfully has the solution approach the line x = π/, but the computation gets slower and slower as t becomes larger. Matlab is choosing smaller and smaller step sizes for larger values of t, which is why it takes it so long to plot the solution. 3. The accuracy of any numerical method in solving a differential equation y = f(t) depends on how strongly the equation depends on the variable y. More precisely, the constants that appear in the error bounds depend on the derivatives of f with respect to y. To see this experimentally, consider the two initial value problems y = y, y() =, y = e t, y() =, You will notice that the two problems have the same solution. For each of the three methods described in this chapter compute approximate solutions to these two initial value problems over the interval [, ] using a step size of h =.. For each method compare the accuracy of the solution to the two problems. Answer: Here are the results of the computations (we ll give 6 decimal places for and and 4 decimal places for ): y = y, y() = y = e t, y() = The value of e to 4 decimal places is We can see that for all of,, and, the approximation from the equation y = e t is more accurate than the approximation from the equation y = y. Of course, the method of approximating e using the differential equation y = e t requires being able to compute e t for several values of t, so it would not be a useful method for approximating e.

12 4. Remember that y(t) = e t is the solution to the initial value problem y = y, y() =. Then e = e, and in MATLAB this is e = exp(). Suppose we try to calculate e approximately by solving the initial value problem, using the methods of this chapter. Use step sizes of the form /n, where n is an integer. For each of Euler s method, the second Runge-Kutta method, and the fourth order Runge-Kutta method, how large does n have to to get an approximation eapp which satisfies eapp e 3? Answer: Since we expect that the fourth order Runge-Kutta will work the best, we will start with it. We use to approximate e using step size and /: >> f = inline('y', 't', 'y') >> [t, y] = (f, [, ],, ); y(size(y,)) >> [t, y] = (f, [, ],, /); y(size(y,)) Using step size h =, we get y().783, so that eapp e =.99, which is not less than.. Using step size h = /, we get y().773, so that eapp e =.9356, which is less than.. Thus, for it works to use n =. Now, we try. We expect this to take larger n. We will create a chart for n =,, 3,.... The relevant information for each n is eapp and eapp e. We will keep increasing n until we find an n such that eapp e <.. n eapp eapp e Thus, for we see that it works to use n =.

13 For, we expect n to be larger and it doesn t seem feasible to compute eapp starting with n = till we get an n that works. Instead, you can jump around. Try n = (which is not large enough), then try n = 5, which is large enough. Then, try some values in between, trying to narrow in on the smallest n that works. You should get that n = 359 is the smallest n that works. Thus, our answer is: For, n must be at least 359. For, n must be at least. For, n must be at least. NSS 4.9. Find a general solution to the differential equation using the method of variation of parameters. y + 4y + 4y = e x ln x Answer: The corresponding homogeneous differential equation is y + 4y + 4y =, with auxiliary equation r + 4r + 4 =, with solution r = as a double root. Thus, the solutions to the homogeneous differential equation are y h = c e x + c xe x. We let y = e x and y = xe x. Then y = e x and y = e x xe x. We want to find a particular solution of the form We get the equations: y p = v e x + v xe x v e x + v xe x = v ( e x ) + v (e x xe x ) = e x ln x We solve the equations for v and v using Cramer s Rule: xe x e x ln x e x xe x v xe = 4x ln x e x xe x = = x ln x e 4x xe 4x + xe 4x e x e x xe x e x e x e x ln x v = e x xe x = e 4x ln x = ln x e 4x e x e x xe x Thus: v = x ln x dx = x ln x + 4 x + C 3

14 v = ln x dx = x ln x x + C Since we just want to find one particular solutions, we can ignore the C and C. Thus: y p = v y +v y = ( x ln x + 4 ) ( x e x +(x ln x x) xe x = x e x ln x 3 ) 4 Thus, the general solution is y = c e x + c xe x + x e x ( ln x 3 4 ) 4. Find a general solution to the differential equation. y + y = tan x + e 3x Answer: We can use the method of undetermined coefficients to solve the differential equation y + y = e 3x, then we will only need to use variation of parameters on y + y = tan x. The auxiliary equation is r + =, which has solutions r = ±i. Thus, the corresponding homogeneous differential equation has solutions y h = c cos x + c sin x. For the differential equation y + y = e 3x, we expect a particular solution of the form y p = Ae 3x + B. Then, y p = 3Ae 3x and y p = 9Ae 3x. Plugging into the differential equation, we get 9Ae 3x + Ae 3x + B = e 3x Thus: Ae 3x + B = e 3x Thus, A = / and B =, so a particular solution is y p = (/)e 3x. Now, we need to use variation of parameters to solve the differential equation y +y = tan x. We have y = cos x and y = sin x. Thus, y = sin x and y = cos x. We want to find a particular solution of the form We get the equations: y p = v cos x + v sin x v cos x + v sin x = v ( sin x) + v cos x = tan x We solve the equations for v and v using Cramer s Rule: sin x tan x cos x v sin x tan x = cos x sin x = cos x + sin = sin x tan x x sin x cos x 4

15 Thus: v = v = cos x sin x tan x cos x sin x sin x = cos x tan x = sin x cos x sin x tan x dx = sin x ln sec x + tan x + C v = sin x dx = cos x + C Since we just want to find one particular solutions, we can ignore the C and C. Thus, a particular solution to the differential equation y + y = tan x is y p = v y + v y = (sin x ln sec x + tan x ) cos x cos x sin x = cos x ln sec x + tan x To get a particular solution to the original differential equation, we combine the two particular solutions: Thus, the general solution is y p = (/)e 3x cos x ln sec x + tan x y = c cos x + c sin x + (/)e 3x cos x ln sec x + tan x 8. Find a general solution to the differential equation. Answer: y + 5y + 6y = 8x We can use the method of undetermined coefficients to solve this differential equation. The auxiliary equation is r + 5r + 6 =, which has solutions r =, 3. Thus, the homogeneous equation has solutions y h = c e x + c e 3x. We expect a particular solution of the form y p = Ax + Bx + C. Then, y p = Ax + B and y p = A. Plugging these into the differential equation, we get: (A) + 5(Ax + B) + 6(Ax + Bx + C) = 8x Combining x, x, and constant terms, we get: 6Ax + (A + 6B)x + (A + 5B + 6C) = 8x Thus, 6A = 8, A + 6B =, and A + 5B + C =. Solving, we get that A = 3, B = 5, and C = 9/6. Thus, the particular solution is y p = 3x 5x + 9/6 5

16 Thus, the general solution is y = c e x + c e 3x + 3x 5x + 9/6. The Bessel equation of order one half, ( x y + xy + x ) y =, x > 4 has two linearly independent solutions, y (x) = x / cos x, y (x) = x / sin x Find a general solution to the nonhomogeneous equation ( x y + xy + x ) y = x 5/, x > 4 Answer: First, we should divide the differential equation by x to put it in the appropriate form: ( y + y x + ) y = x, x > 4x Since y = x / cos x and y = x / sin x, we have y = cos x x x sin x x y = sin x x x + cos x x We want to find a particular solution of the form We get the equations: y p = v x / cos x + v x / sin x v y + v y = v y + v y = x To simplify later calculations, we can compute the Wronskian of y and y : cos x sin x x x W [y, y ] = cos x x x sin x sin x x x x + cos x x = cos x x cos x sin x x + cos x sin x x + sin x x = cos x + sin x x = x 6

17 We solve the equations for v and v using Cramer s Rule: x / sin x x y v = = sin x = x sin x W [y, y ] /x x / cos x y v x = = cos x W [y, y ] /x = x cos x Thus: v = x sin x dx = sin x + x cos x + C v = x cos x dx = cos x + x sin x + C Since we just want to find one particular solutions, we can ignore the C and C. Thus: y p = v y + v y = ( sin x + x cos x) x / cos x + (cos x + x sin x) x / sin x = x cos x + x sin x = x Thus, the general solution is y = c x / cos x + c x / sin x + x 4. Find a general solution to the differential equation given that y and y are linearly independent solutions to the corresponding homogeneous equation for x >. x y 4xy + 6y = x 3 + y = x, y = x 3 Answer: This is a Cauchy-Euler differential equation, so we can use the Cauchy- Euler method along with the method of undetermined coefficients. We use the substitution x = e t, then x = e t x dy dx = dy dt x d y = d y dx dt dy dt This gives us the following differential equation: d y dt dy dt 4dy dt + 6y = e3t + 7

18 This becomes: d y dt 5dy dt + 6y = e3t + This differential equation has auxiliary equation r 5r+6 = with solutions r =, 3. Thus, the corresponding homogeneous equation has solutions y h = c e t + c e 3t (in terms of x, this is y h = c x + c x 3, which agrees with the two homogeneous solutions given in the problem). Since e 3t is a solution to the homogeneous differential equation, we expect a particular solution of the form y p = Ate 3t + B. Thus, y p = 3Ate 3t + Ae 3t and y p = 9Ate 3t + 6Ae 3t. Plugging into the differential equation, we get ( 9Ate 3t + 6Ae 3t) 5 ( 3Ate 3t + Ae 3t) + 6 ( Ate 3t + B ) = e 3t + Thus: (9A 5A + 6A) te 3t + (6A 5A)e 3t + 6B = e 3t + Thus, we get that A = and B = /6. Thus, a particular solution is y p = te 3t + /6 Since t = ln x, this is y p = ln xe 3 ln x + /6 = x 3 ln x + /6 Thus, the general solution to the original equation is y = c x + c x 3 + x 3 ln x + /6 8. Use the method of variation of parameters to show that y(x) = c cos x + c sin x + is a general solution to the differential equation x y + y = f(x) where f(x) is a continuous function on (, ). identity sin(x s) = sin x cos s sin s cos x.] f(s) sin(x s) ds [Hint: Use the trigonometric Answer: The corresponding homogeneous differential equation is y + y =, with auxiliary equation r + =, with solution r = ±i. Thus, the solutions to the homogeneous differential equation are y h = c cos x + c sin x. We let y = cos x and y = sin x. Then y = sin x and y = cos x. We want to find a particular solution of the form y p = v cos x + v sin x 8

19 We get the equations: v cos x + v sin x = v ( sin x) + v (cos x) = f(x) We solve the equations for v and v using Cramer s Rule: sin x f(x) cos x v f(x) sin x = cos x sin x = cos x + sin = f(x) sin x x sin x cos x cos x sin x f(x) v = cos x sin x = f(x) cos x sin x cos x Thus: v = f(x) sin x dx v = f(x) cos x dx Since we just want to find one particular solutions, we can choose these to be following integrals: Thus: v = v = y p = v y + v y = cos x = = x x x x x f(s) sin s ds f(s) cos s ds f(s) sin s ds + sin x x f(s) cos s ds ( f(s) sin s cos x + f(s) cos s sin x) ds f(s) (sin x cos s cos x sin s) ds Now, we use the trig identity from the hint, to get Thus, the general solution is y p = x f(s) sin(x s) ds y = c cos x + c sin x + x f(s) sin(x s) ds 9

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