Chapter 4 Notes and Examples

These notes are just the sketchy highlights.  They do not include all the details, proofs, and examples done in class.  They are meant to give you an overview of what was covered in class.  

How can we solve a higher order, constant coefficient, homogeneous differential equation?  By this is meant an equation of the form

a_ny⁽ⁿ⁾ + a_n-1y^(n-1) + . . . + a₁y' + a₀y = 0.      (A)

Some clever soul many years ago figured out that if we seek a solution of the form y = e^mx we will be able to find solutions to equation (A) by solving the following polynomial equation.

a_nmⁿ + a_n-1m^n-1 + . . . + a₁m + a₀ = 0.      (B)

Why this is true is indicated in the box above and was demonstrated in class along with specific examples.  Maple could be used (Maple worksheet example--multiply, factor polynomials) in doing the algebra involved in finding the zeroes of the auxiliary polynomial and solving the auxiliary equation (B).  In class we went into some detail concerning what it would take to find the general solution to (A).  Ultimately we found that if the auxiliary equation (B) had n distinct roots, m₁, m₂, . . ., m_n then each function y_i = e^mi^x for i = 1, 2, . . ., n would be a solution to (A).  We also discussed why any linear combination of solutions y_i would also be a solution to (A) and in fact concluded that if equation (B) had n distinct roots we could form the general solution to (A) under the assumption that the n solutions of the form y_i = e^mi^x for i = 1, 2, . . ., n are linearly independent.  This is summarized in the box below.

At this point we only informally discussed the meaning of linear independence.  Our conclusion was that we could easily solve the following equation.

y⁽⁴⁾ + 4y⁽³⁾ - 7y" - 22y' + 24y = 0        (C)

because the auxiliary equation, m⁴ + 4m³ - 7m² - 22m + 24 = 0, is easily solved and has solutions m=-4, m=-3, m=1, and m=2.  The general solution to the equation is

y = c₁e^-4x + c₂e^-3x + c₃e^x + c₄e^2x.      (D)

We discussed the need to have n arbitrary constants in forming the general solution to an nth order, constant coefficient, homogeneous differential equation and that we would not actually have n arbitrary constants in (D) above if the solution functions being used were not linearly independent.

We then considered what we could do if (B) did not have n different solutions, i.e., if some of the zeroes of the related polynomial had multiplicity greater than 1.  At this point I introduced the notion of a constant coefficient differential operator, how one would apply such an operator,  and some of the properties of such an operator (such as the fact that it could be factored and the different factors (operators) could be applied in any order and still produce the same result).  For example, in operator form, equation (C) could be written as

(D⁴ + 4D³ - 7D² - 22D + 24)y = 0    or      (D-2)(D-1)(D+3)(D+4)y = 0    

or equivalently as     (D-1)(D-2)(D+4)(D+3)y = 0

where D means differentiation with respect to the independent variable, D² means DD (differentiate and then differentiate again, second derivative) etc.

We looked at applying the operator D-a to the function e^axf(x) where f(x) is a differentiable function and saw that

(D-a)(e^axf(x)) = e^axf '(x) +ae^axf(x) -ae^axf(x) = e^axf '(x)

and then were easily able to see that if the function f is n times differentiable then

(D-a)ⁿ(e^axf(x)) = e^axf ⁽ⁿ⁾(x).

This implies, for example, that if we look at the third order differential equation (D-a)³y = 0 we could find as solutions any function of the form e^axf(x) where the third derivative of f(x) equals zero.  Thus solutions to (D-a)³y = 0 would include y = 1*e^ax, y = x*e^ax, and y = x²*e^ax.  It was noted that we would soon show that these three functions are linearly independent and that the general solution to (D-a)³y = 0 is

y = c₁e^ax + c₂xe^ax + c₃x²e^ax.

We discussed the fact that if the roots of an auxiliary equation have multiplicities greater than one we will be able to find additional solutions (that will be linearly independent) to the corresponding differential equation by multiplying solutions associated with the roots of multiplicity greater than one by the appropriate powers of x.

We next looked at the complication that will arise if some of the roots of the auxiliary equation are not real.  In particular we looked at an equation of the form

ay" + by' + cy = 0      (E) 

where the solutions to am² + bm + c = 0 take the form m = p - iq and m = p + iq where i is the imaginary number sqrt(-1).  In this case it was shown that Euler's formula (Zill, page 134) allows us to transform the general solution to (E) from

y = c₁e^(p-iq)x + c₂e^(p+iq)x     to     y = e^px(c₃cos(qx) + c₄sin(qx))   and we could go back to calling the arbitrary constants c₁ and c₂.  Two (linearly independent) solutions will be y₁ = e^pxcos(qx)   and   y₂ = e^pxsin(qx).  This form will be much more useful when solving for the arbitrary constants when conditions are attached.

DE Example 1

y" + 2y' + 5y = 0      The roots of the auxiliary equation are -1-2i and -1+2i so with p=-1 and q=2 the general solution will be

y = e^-x(c₁cos(2x) + c₂sin(2x))   and two (linearly independent) solutions will be   y₁ = e^-xcos(2x)   and   y₂ = e^-xsin(2x).

In operator form the equation in this example would be written as (D² + 2D +5)y = 0.  

Could we solve the following equation?

(D² + 2D +5)²y = 0

We can.  This equation will also have solutions y₁ = e^-xcos(2x)   and   y₂ = e^-xsin(2x),   and additionally, due to the multiplicities of the zeroes of the auxiliary equation, will have solutions y₃ = xe^-xcos(2x)   and   y₄ = xe^-xsin(2x).  Some of the details were presented in class and it was noted that in this type of case we could multiply solutions related to zeroes of the auxiliary equation with multiplicities greater than two by appropriate powers of x to produce additional (linearly independent) solutions.

For example, one solution to (D² + 2D +5)⁴y = 0 would be y = x³e^-xsin(2x).

Linear Dependence and Linear Independence

Functions y₁, y₂, . . ., y_n are said to be linearly independent if   c₁y₁ + c₂ y₂ + . . .+ c_ny_n = 0 only if  c₁ = c₂  = . . .= c_n= 0.  Otherwise the functions are said to be linearly dependent.  If the functions are linearly dependent then at least one of the constants c_i above would not be zero (say c₁) and we could solve for y₁ in terms of the other functions y_i (as shown in class).

Example

Use the definition above to prove that the following functions are linearly dependent.

y₁ = 2 + x,     y₂ = 3 - x²,     y₃ = 7 + 2x - x²

Since 2*y₁ + 1*y₂ - 1*y₃ = 0 the functions must be linearly dependent (not all c_i = 0).  If you could not figure out in your head values for the constants (not all zero) that would cause the needed linear combination of the functions to be identically (for all x in this case) equal to zero you could form and solve a system of equations as shown below.  In this example that system would have an infinite number of solutions one of which would have been (2,1,-1).  It was also shown in class that if the only solution to this system was (0,0,0) then the functions would be linearly independent.

If the functions had been  y₁ = 2 + x,     y₂ = 2 - x²,     y₃ = 7 + 2x - x²

then we would get the following result:

Suppose two functions, y₁ and y₂ are linearly dependent.  This means that y₂ = cy₁ for some constant c.  In this case the determinant 

det [[y₁,y₂],[y₁',y₂']] was shown in class to be zero.  This determinant is called the Wronskian (Zill, page 145) of the two functions.

After defining the Wronskian for n functions, it was shown in class using properties of determinants that the Wronskian of n linearly dependent functions would be zero.  This is logically equivalent to saying that if the Wronskian of n functions is not zero then the functions must be linearly independent. 

Example

Use the Wronskian to show that the functions e^x, e^2x, and e^3x are linearly independent.  As shown in class and below, the determinant that would be the Wronskian of these three functions will evaluate to 2e^6x, which is not zero, so the functions must be linearly independent.

Click here to see a Maple worksheet showing commands for computing the Wronskian of three functions.

Extra Credit

Use the Wronskian to show that the functions e^ax, e^bx, and e^cx are linearly independent if the constants a, b, and c are three different numbers.

It was shown in class that the Wronskian for the functions defined by y = 1, x, x², . . ., xⁿ would not equal zero and therefore these functions are linearly independent.  This was used to show that the functions defined by y = e^ax, xe^ax, . . ., xⁿe^ax would be linearly independent. 

DE Example 2--Solving a Homogeneous Equation Solve     D2(D+2)(D+1)y = 0     y(0) = 0, y'(0) = 4, y"(0) = -6, y'''(0) = 14 The general solution is    y = c1 + c2x + c3e-2x + c4e-x. y' = c2 - 2c3e-2x - c4e-x,     y" = 4c3e-2x + c4e-x,     y''' = -8c3e-2x - c4e-x Substituting in the initial conditions yields the following system of equations to be solved: 0 = c1 + c3 + c4     4 = c2 - 2c3 - c4,     -6 = 4c3 + c4,     14 = -8c3 - c4 The solution to this system is c1 = 0, c2 = 2, c3 = -2, c4 = 2 so the solution to the differential equation satisfying the initial conditions is y = 2x - 2e-2x + 2e-x.  Maple Worksheet--substituting in initial conditions and solving the system. The graph of the solution is shown on the right in red along with the graph of y = 2x in blue.

Nonhomogeneous equations

How can we solve a higher order, constant coefficient, linear, nonhomogeneous differential equation?  By this is meant an equation of the form

a_ny⁽ⁿ⁾ + a_n-1y^(n-1) + . . . + a₁y' + a₀y = f(x).      or equivalently, in operator form,    

(1)     U(D)y = f(x)     where       U(D) = a_nDⁿ + a_n-1D^n-1 + . . . + a₁D + a₀

U(D) takes the form of a polynomial in D.

You can find the general solution to equation (1) if you can find any solution to (1) and can solve U(D)y = 0.  This is because if y_c (called the complementary function) is the general solution to U(D)y = 0 and y_p is any (particular) solution to (1) then y = y_c + y_p will be the general solution to (1).  It will contain the required number of arbitrary constants and since U(D)(y_c + y_p) = U(D)y_c + U(D)y_p = 0 + f(x) = f(x) it will represent solutions to (1). 

Thus if I can find y_c, the solution to the corresponding homogeneous equation, and find any y_p, any particular solution, I will have the general solution.

DE Example 3

I want to solve    y" - 2y' - 8y = 2e^x - e^-x.      (12)

I solve m² - 2m - 8 = 0 and find the solutions m = -2 and m = 4.  Thus y_c = c₁e^-2x + c₂e^4x.  At this point I am going to take a guess at what y_p will look like.  I am going to "guess" y_p = Ae^x + Be^-x.  It might seem like a good guess considering what the function f(x) (function on the right side of the equal sign above) is in this example.  Substituting this guess for y_p and its first and second derivatives into (12) and combining like terms on the left yields

-9Ae^x - 5Be^-x = 2e^x - e^-x.

Equating coefficients of like terms on each side of the equal sign in the equation above leads to -9A = 2 and -5B = -1 so we need A = -2/9 and B = 1/5 in order that y_p = Ae^x + Be^-x will be a solution to (12).  Thus y_p = (-2/9)e^x + (1/5)e^-x and the general solution to (12) is

y = c₁e^-2x + c₂e^4x + (-2/9)e^x + (1/5)e^-x.

Here is the solution satisfying some initial conditions.

Maple worksheet solution with initial conditions

If the equation I wanted to solve was    y" - 2y' - 8y = 2e^4x - e^-2x    the form to use for y_p would have been less obvious and if I wanted to solve

y" - 2y' - 8y = 2e^4x + cos(2x) - (3x)sin(x)     the form to use for y_p would have been much less obvious. 

If I wanted to solve    y" - 2y' - 8y = [2e^4x + cos(2x)] / [(3x)sin(x)]    I might never figure out a form for y_p and in fact might never be able to solve the equation analytically.

What Can I Solve?

It turns out that one type of equation of the form

(1)     U(D)y = f(x)

that we can solve is the type where we can find a constant coefficient, homogeneous differential equation for which f(x) is a solution.  If we can then we do the following:

To solve (1) first solve the corresponding homogeneous equation, U(D)y = 0, to find y_c.  Next find the simplest (to make it easier) differential operator U^*(D) such that f(x) is a solution to

U^*(D)y = 0.      (2)

For example, if f(x) = 2e^x - e^-x then U^*(D) = (D-1)(D+1) since to solve

(D-1)(D+1)y = 0         (3)

our auxiliary equation would be (m-1)(m+1) = 0 which has solutions m=1 and m=-1 so the general solution to (3) would be y = c₁e^x + c₂e^-x and 2e^x - e^-x is a member of this family of functions (the member where c₁ = 2 and c₂ = -1).

Back to the general procedure.  After finding the differential operator U^*(D), you then form the homogeneous equation

U^*(D)U(D)y = 0.         (4)

If y_p is a solution to (1) it must be a solution to (4) since

U^*(D)U(D)y_p = U^*(D)f(x) = 0.

Thus y_p must take the form of a solution to (4).  We can represent the general solution to (4) as y = y_c + y_q where y_q is the part of the general solution to (4) that is not included in y_c.  Since U(D)y_c = 0 we are left with y_p being a member of the family of functions y_q.  Hopefully an example will make this clearer.

DE Example 4

Solve     y" - 2y' - 8y = 3e^x - e^-2x.       (5)

y_c = c₁e^-2x + c₂e^4x as seen in an earlier example and U(D) = (D+2)(D-4).

The simplest constant coefficient, homogeneous differential equation for which y = 3e^x - e^-2x would be a solution is 

(D-1)(D+2)y = 0. 

Thus U^*(D) = (D-1)(D+2) and the equation in this example corresponding to (4) is

(D-1)(D+2)(D-4)(D+2)y = 0     or equivalently      (D-1)(D-4)(D+2)²y = 0     whose general solution is

y = c₁e^x + c₂e^4x + c₃e^-2x + c₄xe^-2x.

Since y_c = c₃e^-2x + c₂e^4x    we are left with y_q = c₁e^x + c₄xe^-2x  which means that our form for y_p is y_p = Ae^x + Bxe^-2x.  We now substitute this y_p for y in (5) to try to find A and B.  Doing so yields    -9Ae^x - 6Be^-2x = 3e^x - e^-2x.       Equating coefficients of like terms leads to A = -1/3 and B = 1/6.

Thus y_p = (-1/3)e^x + (1/6)xe^-2x    and the general solution to equation (5) is

y = c₁e^-2x + c₂e^4x + (-1/3)e^x + (1/6)xe^-2x.

Here is a PowerPoint presentation on finding the form for a particular solution (same presentation with audio using Flash). 

DE Example 5

Solve     D²(D-1)y = 3e^x + sin(x)      (6)

y_c = c₁ + c₂x + c₃e^x

In this case the form you would need for y_p (as shown in class) would be

y_p = Axe^x + Bsin(x) + Ccos(x)

Substituting y_p above in for y in (6) leads to

Ae^x + (-B+C)cos(x) + (B+C)sin(x) = 3e^x + sin(x).       (7)

To find A, B, and C for y_p to be a particular solution to (6) we need to equate coefficients of like terms in (7) and solve the resulting system of equations shown below.

A = 3,     -B + C = 0,     B + C = 1

Solving this system yields    y_p = 3xe^x + (1/2)sin(x) + (1/2)cos(x)    and the general solution to (6) is

y = c₁ + c₂x + c₃e^x + 3xe^x + (1/2)sin(x) + (1/2)cos(x).

Example

Solve    (D²-D)y = 2 - 2x

y_c = c₁ + c₂e^x      The form for y_p is    y_p = Ax + Bx².

Note the multiplicity in the U(D) and U^*(D) operators.  The common mistake is to try to use y_p = A + Bx.  In this example it turns out that A = 0 and B = 1 so the general solution is

y = c₁ + c₂e^x + x²

Examples for Practice

See if you could figure out the U^*(D) operator for each of the following functions.

1.  x² - 5sin(3x)       2.  2e^xcos(3x)      3.  xe^-xsin(2x) + 4cos(3x)     4.  2cosh(2x) - sinh(2x)     5.  x²sin(x)

6.  x²sin(x) + xcos(x)     7.  x² + 3e^-4x     8.  4 - 5x + x² + 3e^-4x     9.  3x - sin(x) + 4cos(x) - e^-2x

Answers

1.  D³(D²+9)     2.  [(D-1)²+9]     3.  [(D+1)²+4]²[D²+9]     4.  D²-4  (Think about the definition of these hyperbolic functions.)

5.  (D²+1)³     6.  Same as 5     7.  D³(D+4)     8.  Same as 7     9.  D²(D²+1)(D+2)

Suppose the differential equation involving the function in number 9 above had been

(D³+D)y = 3x - sin(x) + 4cos(x) - e^-2x.    

In this case, due to the multiplicities involved, the form for y_p would have been

y_p = Ax + Bx² + Cxsin(x) + Excos(x) + Fe^-2x

Make sure you could have figured this out yourself.

Extra Credit

In class I solved the following problem:

(D²+1)y = x³,     y(0) = 0,  y(pi) = 0

I showed that there is no solution to the differential equation satisfying the given conditions.

You solve the following problem and show that there are an infinite number of solutions.

(D²+1)y = 2cos(x),     y(0) = 0,  y(pi) = 0

Explain why the fact that each of these problems does not have one unique solution does not contradict the Existence of a Unique Solution Theorem 4.1.1 given on page 117 of your text (Zill, 10th edition).

Here is an example applying undetermined coefficients to a first order linear differential equation.

DE Example 6

Graph of the Solution

Maple Worksheet for DE Examples 1 - 6

DE Example 7

Maple commands to find A, B, C

A Damped Forced Vibration Example In this example I will look at a spring mass system with damping and a forcing function that yields the initial value problem shown below.  The initial position and initial velocity are each zero. Maple Worksheet	Excel Version of the Graph
	The solution is graphed above in orange.  The steady-state solution is graphed in green.  The transient solution is graphed in blue.  Click here (QT) to see an animated graph of y' versus y as t goes from zero to six pi.  y' is along the vertical axis.  Click here for an animation that includes the changing value of t.  Click here to link to a wonderful applet dealing with damped forced vibrations.  I will demo this in class and explain more.  Here is a short video of me using the applet.  Entering m = 1, b = 3, k = 2, F = 2, and W = 1 in the applet will duplicate the problem on the left.  Try entering m = 1, b = 0, k = 1, F = 1, and W = 1 and see what happens.  This relates to the spring mass Exam II practice problem.  Try setting m = 1, b = 0, k = 2, F = 2, and W = 1.  Try setting b and F equal to zero (and nothing else zero).  What is happening in this case?  Set F equal to zero (and nothing else zero).  What is happening in this case?  EC1:  Give the parametric representation of the graph of y' versus y in my animation above.
	EC2:  Run the applet for m = 1, b = 0.1, k = 1, F = 1, W = 1, x(0) = 0, and x'(0) = 0.  Find the solution, transient solution, and steady-state solution analytically.  The solution is graphed above in orange.  The steady-state solution is graphed in green.  The transient solution is graphed in blue.  After doing this, set b = 0, leave the other values alone, and run the applet.  What happens?   EC3:  Prove that the graph of y' versus y for the steady-state part of the solution shown on the left is a circle whose radius is the square root of 2/5.   Here is an Animation of solutions to       y'' + k2y = cos(t),  y(0) = 0,  y'(0) = 0 with m = 1, b = 0, k = 1 in red and k varying from 0 to 2 in blue, F = 1, W = 1.  The animation also works for       y'' + y = cos(Wt),  y(0) = 0,  y'(0) = 0 with W = 1 in red and W varying from 0 to 2 in blue. Here is a similar animation QT over a larger t interval and in this animation W (or k) varies from 0 to 4 in blue. Analytical Solutions

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        Lane Vosbury, Mathematics, Seminole State College   email:  vosburyl@seminolestate.edu

        This page was last updated on 08/21/14          Copyright 2002          webstats