Problem set #6
1. (5.7[1]) Solve the initial value problem
x’ = x with x(0.1) =
Ans:
x = 0.5556e3t + 0.2763e-t
2. (5.91) Convert the third-order differential equation
y’’’(x) 5y’’(x) y’(x) + 5y(x) = 0
to a first-order system and find the eigenvalues and eigenvectors of the system matrix.
Ans:
x’ = x
The eigenvalues are 1, 1, 5. The corresponding eigenvectors are
, , and
5.10 Solve the equation
(t) + 3(i) + 2y(t) = U(t),
in which U(t) is the unit step function
U(t) =
Convert the equation to a first-order matrix equation. Then find the fundamental matrix and use the method of variation of parameters. Assume the system is initially at rest.
Solution
The second-order equation becomes
= + with x(0) =
This system has the form = Ax + f. The eigenvalues of A are 1, 2, and two eigenvectors are u(1) = [1, 1]T, and u(2) = [1, 2]T. The solution is
x = (t)c + xp
(t) is the fundamental matrix. (t) =
Assuming a particular solution of the original equation xp = (t)u leads to the relationship (t) = f.
= -1(t)f = = u =
xp = (t)u = =
x = c + x(0) = = + c = [1, 1/2]T.
x = +
> [x1,x2]=dsolve('Dx=y,Dy=-2*x-3*y+1','x(0)=0,y(0)=0')
x1 =
1/2*exp(-2*t)-exp(-t)+1/2
x2 =
exp(-t)-exp(-2*t)
5.12 Consider the second-order differential equation
(t) + 2(i) + 02y(t) = 0
which might represent a series RLC circuit or a mechanical system with a mass, spring, and dashpot.
a. What is the significance of the case when the characteristic equation yields repeated eigenvalues?
b. Write the homogeneous solution in the case of a double eigenvalue.
Solution
a. The characteristic equation 2 + 2 + 02 = 0 yields the eigenvalues
1,2 =
When = 0, the characteristic equation yields repeated and real eigenvalues. The response is said to be critically damped.
b. The critically damped response for the homogenous equation is
y(t) = e-t(c1 + c2t)
5.16 Write an M-file to solve the equation (t) + 3(i) + 2y(t) = 1. The system is initially at rest.
Solution
% P5_16.M Solve the equation D2y+3*Dy+2*y = 1
% y(0)=0,y'(0)=0
%
y1=dsolve('D2y+3*Dy+2*y=1','y(0)=0,Dy(0)=0')
%
tf=1.0;
clf
ezplot(y1,[0,tf]);
y1 =
1/2*exp(-2*t)-exp(-t)+1/2
6. Solve the following equation with step size h = 0.2
= x - 2y, at x = 1, y = 1.4
1. Using Euler methody(1.4) = _0.864___
2. Using modified Euler method: yn+1 = yn+ .5h[f(xn,yn)+f(xn+1,yn+h*f(xn,yn))]
y(1.2) = __1.132___
3. Using Midpoint method: yn+1 = yn+ k2 where k1 = h*f(xn,yn) and
k2 = h*f(xn+.5h,yn+0.5k1)y(1.2) = ___1.132___
4. Using Heun’s method: yn+1 = yn+ (k1 + k2) where k1 = h*f(xn,yn) and
k2 = h*f(xn+h, yn+k1)y(1.2) = ___1.132___
7. Solve the following equations with step size h = 0.2
= - y2 + xy1, y1(1) = 2; = y2 + x, y2(1) = -1
1. Using Euler method, the value of y1 at x = 1.2 is ____2.6____
2.Using modified Euler method, the value of y1 at x = 1.2 is ____2.712____
3. Using Midpoint method, the value of y1 at x = 1.2 is ____2.706__
8. Determine the stiffness ratio for the following set of equations
= -600y1 + 590y2; = 590y1 – 600y2SR = ___119____
10. Solve = y using the fourth-order Runge-Kutta method with y(0) = 1 using h = .5 to evaluate
y(1). Solve by hand and verify with MATLAB ode45.
Solution
yn+1 = [1 + h + (h)2 + (h)3 + (h)4] yn = [1 + h + (h)2 + (h)3 + (h)4] yn
y(0.5) = [1 + .5 + (.5)2 + (.5)3 + (.5)4](1) = 1.6484
y(1.0) = [1 + .5 + (.5)2 + (.5)3 + (.5)4]( 1.6484) = 2.7173
[1]Advanced Engineering Mathematics with MATLAB, 2e by Thomas Harman