Section 2.1 Introduction: Second-Order Linear Equations

Second Order Differential Equations: F(x, y, y′, y″) = 0 2 nd Order Linear Differential Equations: A(x) y″ + B(x) y′ + C(x) y = f (x) often these will be in the form y″ + P(x) y′ + Q(x) y = f (x) n th Order Linear Differential Equations

Homogeneous Linear Differential Equations: A(x) y″ + B(x) y′ + C(x) y = 0 Solutions to homogeneous linear Differential Equations form vector spaces!

Theorem Let y 1 and y 2 be solutions to y″ + P(x) y′ + Q(x) y = 0. Then functions of the form f (x) = c 1 y 1 + c 2 y 2 (with c 1 and c 2 being any constants) are also solutions.

General Solutions Initial Conditions

Theorem Suppose the P(x) and Q(x) are continuous on an interval J containing x o. Then the initial value problem y″ + P(x) y′ + Q(x) y = 0, y′(x o ) = a, y(x o ) = b has a unique solution on J.

Ex. 1 Solve y″ + y = 0,

Suppose we've found two linearly independent solutions, y 1 and y 2, to the initial value problem y″ + P(x) y′ + Q(x) y = 0, y′(x o ) = a, y(x o ) = b. How do we find the general solution to this differential equation?

Recall: Two functions f (x) and g(x) are said to be linearly independent if one function cannot be written as a multiple of the other (or equivalently, f (x) and g(x) are linearly independent if the equation c 1 f (x) + c 2 g(x) = 0 has only a solution of c 1 = c 2 = 0).

Ex. 2 Show that y 1 = sin(x) and y 2 = 3sin(x) are both solutions to y″ + y = 0, but we cannot use these to find our particular solution.

Definition: The Wronskian of f (x) and g(x) is defined as:

Ex. 3 Calculate the Wronskian of f (x) = sin(x) and g(x) = x 2.

Theorem: Suppose y 1 and y 2 are solutions of y″ + P(x) y′ + Q(x) y = 0 on an interval J in which P(x) and Q(x) are continuous. (a) If y 1 and y 2 are linearly dependent then W(f, g) = 0 for all x on the interval J. (b) If y 1 and y 2 are linearly independent then W(f, g) ≠ 0 for any x on the interval J.

Ex. 4 (a) What does this theorem imply regarding the solutions y 1 = sin(x) and y 2 = cos(x) to the differential equation y″ + y = 0?

Ex. 4 (b) What does this theorem imply regarding the solutions y 1 = sin(x) and y 2 = 3sin(x) to the differential equation y″ + y = 0?

Theorem: Suppose y 1 and y 2 are linearly independent solutions of y″ + P(x) y′ + Q(x) y = 0 where P(x) and Q(x) are continuous on an interval J. If g(x) is any solution to this differential equation then g(x) = c 1 y 1 + c 2 y 2 for some choices of constants c 1 and c 2.

Big Important Question: How do we determine two linearly independent solutions, y 1 and y 2 ? We shall examine this question for the differential equation ay″ + by′ + cy = 0 (where a, b, and c are constants).

Claim: y = e rx is a solution (for some constant r) to ay″ + by′ + cy = 0. We just have to figure out what value of r will work.

Ex. 5 Solve y″ – 5y′ + 6y = 0, y(0) = 1, y′(0) = –1.

Theorem: If the roots of the characteristic equation ar 2 + br + c = 0 are r 1 and r 2 with r 1 ≠ r 2 then y 1 = _________ and y 2 = _________ are two linearly independent solutions to the differential equation ay″ + by′ + cy = 0. Thus, y = _________________ is the general solution to ay″ + by′ + cy = 0.

What if there is a repeated root of the characteristic equation? Ex. 6 Solve y″ – 8y′ + 16y = 0, y(0) = 5, y′(0) = 21.

Theorem: If r 1 is a repeated root of the characteristic equation ar 2 + br + c = 0 then y 1 = _________ and y 2 = _________ are two linearly independent solutions to the differential equation ay″ + by′ + cy = 0. Thus, y = _________________ is the general solution to ay″ + by′ + cy = 0.

Ex. 6 Solve y″ – 8y′ + 16y = 0, y(0) = 5, y′(0) = 21.

Why does this theorem work? Let's prove that it produces the correct general solution for this past problem. (Introduction to the linear operator D.)

Why does y″ + y = 0 have solutions of y 1 = sin(x) and y 2 = cos(x)???? Our previous theorems couldn't have possibly generated these solutions.

Ex. 7 Use our past two theorems to solve y″ + y = 0.

Recall that e z =

Recall that e z = cos(z) =

Recall that e z = cos(z) = sin(z) =

Recall that e z = cos(z) = sin(z) =

Ex. 8 Solve y″ – 6y′ + 13y = 0

Theorem: If the roots of the characteristic equation ar 2 + br + c = 0 are A ± Bi then y 1 = e Ax sin(Bx) and y 2 = e Ax cos(Bx) are two linearly independent solutions to the differential equation ay″ + by′ + y = 0. Thus, y = c 1 e Ax sin(Bx) + c 2 e Ax cos(Bx) is the general solution to ay″ + by′ + y = 0.

Section 2.2 General Solutions of Linear Equations

Higher order linear differential equations: y (n) + p n–1 (x) y (n–1) + p n–2 (x) y (n–2) + · · · · · + p 2 (x) y″ + p 1 (x) y′ + p 0 (x) y = f (x)

If y 1, y 2,..., y n are linearly independent solutions to the homogeneous diff. eq. y (n) + p n–1 (x) y (n–1) + p n–2 (x) y (n–2) + · · · · · + p 1 (x) y′ + p 0 (x) y = 0, then the general solution to y (n) + p n–1 (x) y (n–1) + · · · · · + p 1 (x) y′ + p 0 (x) y = 0 is y = c 1 y 1 + c 2 y 2 + · · · + c n y n.

To prove this we needed the following definitions and theorems: Theorem: If y 1, y 2,..., y n are solutions to y (n) + p n–1 (x) y (n–1) + · · · · · + p 1 (x) y′ + p 0 (x) y = 0 then so is y = c 1 y 1 + c 2 y 2 + · · · + c n y n Theorem: Suppose y 1, y 2,.., y n are solutions to y (n) + p n–1 (x) y (n–1) + ··· + p 1 (x) y′ + p 0 (x) y = 0 on an interval J where the p i (x) are continuous. (a) If y 1, y 2,..., y n are linearly dependent then W(y 1, y 2,..., y n ) = 0 at each point on the interval J. (b) If y 1, y 2,..., y n are linearly independent then W(y 1, y 2,..., y n ) ≠ 0 at each point on the interval J. Theorem: If y 1, y 2,..., y n are solutions to y (n) + p n–1 (x) y (n–1) + · · · · · + p 1 (x) y′ + p 0 (x) y = 0, then so is y = c 1 y 1 + c 2 y 2 + · · · + c n y n for any choice of constants c 1, c 2,..., c n. Theorem: If y 1, y 2,..., y n are linearly independent solutions to y (n–1) + p n–2 (x) y (n–2) + · · · · · + p 1 (x) y′ + p 0 (x) y = 0, then general solution is y = c 1 y 1 + c 2 y 2 + · · · + c n y n.

Ex. 1 Suppose that the following functions are all particular solutions to y (3) + 3y″ + 4y′ + 12y = 0. We would like to have a set of three linearly independent solutions. If this can be done then which three can we use? If it cannot be done then show that it cannot. y 1 = sin(2x) y 2 = cos(2x) y 3 = 5 sin(2x) + 3 cos(2x) y 4 = sin(x) cos(x) y 5 = cos 2 (x) – sin 2 (x) y 6 = e –3x

Ex. 2 Use the definition of linear dependence to show that the following functions are linearly dependent. f (x) = x 2 + 3x, g(x) = 7x 2, h(x) = 31x 2 + 6x.

The Wronskian of the n functions f 1 (x), f 2 (x),..., f n (x):

Theorem: Suppose y 1, y 2,..., y n are solutions of y (n) + p n–1 (x) y (n–1) + ··· + p 1 (x) y′ + p 0 (x) y = 0 on an interval J in which p n–1 (x), p n–2 (x), · · · p 1 (x), p 0 (x) are continuous. (a) If y 1, y 2,..., y n are linearly dependent then W(y 1, y 2,..., y n ) = 0 for all x on the interval J. (b) If y 1, y 2,..., y n are linearly independent then W(y 1, y 2,..., y n ) ≠ 0 for all x on the interval J.

Ex. 3 (a) y 1 = sin(2x), y 2 = cos(2x), y 3 = e –3x are solutions to y (3) + 3y″ + 4y′ + 12y = 0. Use the Wronskian to determine whether or not they are linearly independent.

Ex. 3 (b) y 1 = sin(2x), y 2 = cos(2x), y 3 = e –3x are solutions to y (3) + 3y″ + 4y′ + 12y = 0. Use the Wronskian to determine whether or not they are linearly independent.

Ex. 3 (c) y 1 = x, y 2 = x ln(x), y 3 = x 2 are solutions to x 3 y (3) – x 2 y″ + 2xy′ – 2y = 0. Use the Wronskian to determine whether or not they are linearly independent.

Suppose we can find three linearly independent solutions to y (3) + 3 y″ + 4y′ + 12y = 0. Are we guaranteed a solution to the initial valued problem with given initial conditions of y(0) = 1, y′(0) = 2, y″(0) = 3? Why??

Theorem: Suppose p n–1 (x), p n–2 (x), · · · p 1 (x), p 0 (x) are continuous on an interval J which contains x o. If we can find n linearly independent solutions to the diff. eq. y (n) + p n–1 (x) y (n–1) + · · · · · + p 1 (x) y′ + p 0 (x) y = 0 then we are guaranteed a solution to the initial value problem of y (n) + p n–1 (x) y (n–1) + · · · · · + p 1 (x) y′ + p 0 (x) y = 0 y(x 0 ) = b 0 y′(x 0 ) = b 1 y″(x 0 ) = b 2 : y (n) (x 0 ) = b n

Theorem: If y 1, y 2,..., y n are linearly independent solutions to y (n) + p n–1 (x) y (n–1) + · · · + p 1 (x) y′ + p 0 (x) y = 0, then general solution is y = c 1 y 1 + c 2 y 2 + · · · + c n y n.

Nonhomogeneous linear Differential Equations: y (n) + p n–1 (x) y (n–1) + · · · · · + p 1 (x) y′ + p 0 (x) y = f (x) We solve a nonhomogeneous differential equation by first examining the corresponding homogeneous differential equation. Review:

Theorem: Suppose that y c = c 1 y 1 + c 2 y 2 + · · · + c n y n is the general solution to the homogeneous differential equation y (n) + p n–1 (x) y (n–1) + ··· + p 1 (x) y′ + p 0 (x) y = 0. Further suppose that y p is any particular solution to the nonhomogeneous differential equation y (n) + p n–1 (x) y (n–1) + · · · · · + p 1 (x) y′ + p 0 (x) y = f (x). Then the general solution to the nonhomogeneous differential equation y (n) + p n–1 (x) y (n–1) + · · · · · + p 1 (x) y′ + p 0 (x) y = f (x) is: y = y p + y c = y p + c 1 y 1 + c 2 y 2 + · · · + c n y n

Ex. 4 (a) Verify that y = ( –1 / 4 ) x is a particular solution to the diff. eq. y″ – 4y = x.

Ex. 4 (b) Verify that y = ( –1 / 4 ) x + e 2x is another particular solution to the diff. eq. y″ – 4y = x.

Ex. 4 (c) What is the general solution to the differential equation y″ – 4y = x ?

Ex. 4 (d) Find a solution to the initial value problem y″ – 4y = x, y(0) = 11, y′(0) = –2.

Ex. 5 (a) y = 2x is one solution to the differential equation y″ – 10y′ + 21y = 42x – 20. Use this to find the general solution.

Ex. 5 (b) y = 2x + 5e 3x is another solution to the differential equation y″ – 10y′ + 21y = 42x – 20. Use this to find the general solution.

Section 2.3 Homogeneous Equations with Constant Coefficients

Find the general solution to y (3) + y″ – 17y′ + 15y = 0

Recall: If r 1 and r 2 are distinct roots of the characteristic equation ar 2 + br + c = 0, then and are linearly independent solutions to the differential equation ay″ + by′ + cy = 0, and thus is the general solution to ay″ + by′ + cy = 0.

Recall: If r 1 and r 2 are distinct roots of the characteristic equation ar 2 + br + c = 0, then and are linearly independent solutions to the differential equation ay″ + by′ + cy = 0, and thus is the general solution to ay″ + by′ + cy = 0. Now: Theorem: If r 1, r 2,..., r k are distinct roots of the characteristic equation a n r n + a n–1 r n–1 + · · · + a 1 r + a 0 = 0, then are linearly independent solutions to the differential equation a n y (n) + a n–1 y (n–1) + · · · + a 1 y′ + a 0 y = 0.

Ex. 1 Find the general solution to y (3) + y″ – 17y′ + 15y = 0.

Ex. 2 Find the particular solution to y (3) + y″ – 17y′ + 15y = 0 y(0) = 10 y′(0) = 8 y″(0) = 50

Ex. 3 Find the general solution to y (5) – 3y (4) + 2y (3) + 2y″ – 3y′ + y = 0.

Theorem: If r 1 is a root of multiplicity k in the characteristic equation a n r n + a n–1 r n–1 + · · · + a 1 r + a 0 = 0, then are linearly independent solutions to the differential equation a n y (n) + a n–1 y (n–1) + · · · + a 1 y′ + a 0 y = 0.

Ex. 4 Find the general solution to y (9) – 25y (8) + 264y (7) – 1520y (6) y (5) – 9984y (4) y (3) – 4096y″ = 0.

Section 2.5 Nonhomogeneous Equations & Undetermined Coefficients

Recall: Theorem: Suppose that y c = c 1 y 1 + c 2 y 2 + · · · + c n y n is the general solution to the homogeneous differential equation y (n) + p n–1 (x) y (n–1) + ··· + p 1 (x) y′ + p 0 (x) y = 0. Further suppose that y p is any particular solution to the nonhomogeneous differential equation y (n) + p n–1 (x) y (n–1) + · · · · · + p 1 (x) y′ + p 0 (x) y = f (x). Then the general solution to the nonhomogeneous differential equation y (n) + p n–1 (x) y (n–1) + · · · · · + p 1 (x) y′ + p 0 (x) y = f (x) is: y = y p + y c = y p + c 1 y 1 + c 2 y 2 + · · · + c n y n

Recall: Ex. 5 (a) y = 2x is one solution to the differential equation y″ – 10y′ + 21y = 42x – 20. Use this to find the general solution. How do we determine y p though?

The Method of Undetermined Coefficients (Using an educated guess to determine y p )

Ex. 1 y″ + 2y′ – 3y = 3e 2x (a) Make a guess on what y p should look like for this differential equation? (What "form" should y p have?)

Ex. 1 y″ + 2y′ – 3y = 3e 2x (b) Determine, specifically, what y p is.

Ex. 1 y″ + 2y′ – 3y = 3e 2x (c) Give the general solution to this nonhomogeneous differential equation.

Ex. 2 y″ + 2y′ – 3y = 30 cos(x) (a) Make a guess on what y p should look like for this differential equation? (What "form" should y p have?)

Ex. 2 y″ + 2y′ – 3y = 30 cos(x) (b) Determine, specifically, what y p is.

Ex. 2 y″ + 2y′ – 3y = 30 cos(x) (c) Give the general solution to this nonhomogeneous differential equation.

Ex. 3 Find the general solution to 2y″ + 4y′ + 7y = x 2

Ex. 4 Find the general solution to y″ – 4y = 2e 2x

Ex. 5 Find the form of y p for the nonhomogeneous differential equation: y″ – 4y = 2xe 2x

Ex. 6 Find the form of y p for the nonhomogeneous differential equation: y (3) + 9y′ = x sin(7x) + x 2 e 2x

Ex. 7 Find the form of y p for the nonhomogeneous differential equation: (D – 2)(D – 5) 3 y = xe 2x + x 2 e 5x + cos(x)

Lemma 1: For any function u(x) we have: (D – r)[u(x)e rx ] = (D [u(x)]) (e rx ) Proof: (D – r)[u(x)e rx ] = D[u(x)e rx ] – r[u(x)e rx ] = D[u(x)] e rx + u(x)D[e rx ] – ru(x)e rx = D[u(x)] e rx + ru(x)e rx – ru(x)e rx = D[u(x)] e rx

Lemma 2: For any function u(x) we have: (D – r) k [u(x)e rx ] = (D k [u(x)]) (e rx )

Question: Find the form of y c for the differential equation (D – r) 3 y = (–3 + 2x)e rx Solution: We compute y c : y c = c 1 e rx + c 2 xe rx + c 3 x 2 e rx Our initial guess for y p : y p = Ae rx + Bxe rx Our adjusted guess for y p : y p = Ax 3 e rx + Bx 4 e rx Why did this second guess work? This is the correct form for y p, but why? We have seen why our first guess at y p won't work. But why does throwing in higher powers of x into these terms give us the correct guess for y p ?

Before we answer this we must first note the following: Note that if we determined the values of A and B we would have our general solution to this nonhomogeneous equation be y = c 1 e rx + c 2 xe rx + c 3 x 2 e rx + Ax 3 e rx + Bx 4 e rx Now let's look at the question again, this time without relying on our theorem from chapter 2 which indicates that throwing in higher powers of x into these terms give us the correct guess for y p.

Question: Find the form of y p for the differential equation (D – r) 3 y = (–3 + 2x)e rx Solution: We transform this difficult nonhomogeneous diff. eq. into an easy homogeneous differential equation:

Question: Find the form of y p for the differential equation (D – r) 3 y = (–3 + 2x)e rx Solution: If y is a solution to (D – r) 3 y = (–3 + 2x)e rx then y is a solution to (D – r) 2 [(D – r) 3 y] = (D – r) 2 [(–3 + 2x)e rx ] then y is a solution to (D – r) 5 y = (D – r) 2 [(–3 + 2x)e rx ] then y is a solution to (D – r) 5 y = D 2 [(–3 + 2x)] e rx then y is a solution to (D – r) 5 y = 0 e rx then y is a solution to (D – r) 5 y = 0 Thus y takes the following form: y = c 1 e rx + c 2 xe rx + c 3 x 2 e rx + c 4 x 3 e rx + c 5 x 4 e rx y = c 1 e rx + c 2 xe rx + c 3 x 2 e rx + Ax 3 e rx + Bx 4 e rx y c y p Since we know that this first part comes from y c, the second part must come from y p. Therefore the form of y p is y p = Ax 3 e rx + Bx 4 e rx.

Ex. 8 Find the general solution to (D – 2) 5 y = e 2x + 3xe 2x

The Method of Variation of Parameters

Try to use the method of undetermined coefficients to solve: y″ + y = tan(x)

Suppose we are given the differential equation y″ + P(x) y′ + Q(x) y = f (x) and we have determined that y c = c 1 y 1 + c 2 y 2. It turns out that y p can be formed from y 1 and y 2. In fact there exists some functions u 1 (x) and u 2 (x) so that y p = u 1 y 1 + u 2 y 2. We are given the differential equation y″ + P(x) y′ + Q(x) y = f (x), and we have determined that y c = c 1 y 1 + c 2 y 2. Suppose y p = u 1 y 1 + u 2 y 2 is a particular solution to the nonhomogeneous equation (where u 1 and u 2 are some functions unknown to us). We must determine what u 1 and u 2 are.

y p = u 1 y 1 + u 2 y 2 y p ′ = (u 1 ′ y 1 + u 1 y 1 ′ ) + (u 2 ′ y 2 + u 2 y 2 ′ ) = (u 1 y 1 ′ + u 2 y 2 ′ ) + (u 1 ′ y 1 + u 2 ′ y 2 ) We shall now impose a restriction of u 1 ′ y 1 + u 2 ′ y 2 = 0 So, y p ′ = u 1 y 1 ′ + u 2 y 2 ′ y p ″ = (u 1 y 1 ′ + u 2 y 2 ′ )′ y p ″ = (u 1 ′ y 1 ′ + u 1 y 1 ″ ) + (u 2 ′ y 2 ′ + u 2 y 2 ″ ) y p ″ = u 1 ′ y 1 ′ + u 2 ′ y 2 ′ + u 1 y 1 ″ + u 2 y 2 ″ y p ″ = u 1 ′ y 1 ′ + u 2 ′ y 2 ′ + u 1 [–P(x) y 1 ′ – Q(x) y 1 ] + u 2 [–P(x) y 2 ′ – Q(x) y 2 ] y p ″ = u 1 ′ y 1 ′ + u 2 ′ y 2 ′ – P(x) [u 1 y 1 ′ + u 2 y 2 ′] – Q(x) [u 1 y 1 + u 2 y 2 ] y p ″ = u 1 ′ y 1 ′ + u 2 ′ y 2 ′ – P(x) [y p ′ ] – Q(x) [y p ] y p ″ + P(x) y p ′ + Q(x) y p = u 1 ′ y 1 ′ + u 2 ′ y 2 ′ f (x) = u 1 ′ y 1 ′ + u 2 ′ y 2 ′

Now we have two conditions placed on u 1 and u 2 : u 1 ′ y 1 + u 2 ′ y 2 = 0 u 1 ′ y 1 ′ + u 2 ′ y 2 ′ = f (x)

Now we have two conditions placed on u 1 and u 2 : u 1 ′ y 1 + u 2 ′ y 2 = 0 u 1 ′ y 1 ′ + u 2 ′ y 2 ′ = f (x) Cramer's rule states that we can solve this for u 1 ′ and u 2 ′ as long as

Now we have two conditions placed on u 1 and u 2 : u 1 ′ y 1 + u 2 ′ y 2 = 0 u 1 ′ y 1 ′ + u 2 ′ y 2 ′ = f (x) Cramer's rule states that we can solve this for u 1 ′ and u 2 ′ as long as Cramer's rule states

Now we have two conditions placed on u 1 and u 2 : u 1 ′ y 1 + u 2 ′ y 2 = 0 u 1 ′ y 1 ′ + u 2 ′ y 2 ′ = f (x) Cramer's rule states that we can solve this for u 1 ′ and u 2 ′ as long as Cramer's rule states

Now we have two conditions placed on u 1 and u 2 : u 1 ′ y 1 + u 2 ′ y 2 = 0 u 1 ′ y 1 ′ + u 2 ′ y 2 ′ = f (x) Cramer's rule states that we can solve this for u 1 ′ and u 2 ′ as long as Cramer's rule states Now, integrate to determine u 1 and u 2.

Ex. 9 Use variation of parameters to solve: y″ + y = tan(x)

Ex. 10 Use variation of parameters to solve: y″ – 4y = 2e 2x

Section 2.4 Mechanical Vibrations

Forces Involved: F S = –kx (k > 0) Restorative Force (Hooke's Law) F R = –cx′ (c > 0) Force due to dashpot

Total Force = F R + F S mx″ = –cx′ – kx mx″ + cx′ + kx = 0 Later we will include external force and get the differential equation mx″ + cx′ + kx = F(t) Terminology: If c = 0 the motion is undamped If c > 0 the motion is damped If F(t) = 0 the motion is free If F(t) ≠ 0 the motion is forced

Free Undamped Motion mx″ + kx = 0 : x(t) = A cos(ω o t) + B sin(ω o t) where A and B are some constants and : x(t) = C cos(ω o t – α) where C and α are some constants and

Free Undamped Motion mx″ + kx = 0 : x(t) = A cos(ω o t) + B sin(ω o t) where A and B are some constants and : x(t) = C cos(ω o t – α) where C and α are some constants and

Free Undamped Motion mx″ + kx = 0 : x(t) = A cos(ω o t) + B sin(ω o t) where A and B are some constants and : x(t) = C cos(ω o t – α) where C and α are some constants and For this last equation we have the following interpretations: C Amplitude ω o (Circular) Frequency αPhase Angle Frequency Period

Ex. 1 A body with mass m = ( 1 / 2 )kg is attached to the end of a spring that is stretched 2m by a force of 100 N. It is set in motion with initial position x o = 0.5m and initial velocity v o = –10m/s. Find the position function of the body as well as the amplitude, frequency, period of oscillation, and phase angle of its motion.

Damped Free Motion mx″ + cx′ + kx = 0

Damped Free Motion mx″ + cx′ + kx = 0 The characteristic equation has roots:

Damped Free Motion mx″ + cx′ + kx = 0 The characteristic equation has roots: Whether these roots are real, complex, or repeated depends on the sign of c 2 – 4km. So, the sign of c 2 – 4km really amounts to whether c =, c

Damped Free Motion mx″ + cx′ + kx = 0 The characteristic equation has roots: Whether these roots are real, complex, or repeated depends on the sign of c 2 – 4km. So, the sign of c 2 – 4km really amounts to whether c =, c Critical damping: (results in repeated roots)

Damped Free Motion mx″ + cx′ + kx = 0 The characteristic equation has roots: Whether these roots are real, complex, or repeated depends on the sign of c 2 – 4km. So, the sign of c 2 – 4km really amounts to whether c =, c Overdamping: (results in distinct real roots)

Damped Free Motion mx″ + cx′ + kx = 0 The characteristic equation has roots: Whether these roots are real, complex, or repeated depends on the sign of c 2 – 4km. So, the sign of c 2 – 4km really amounts to whether c =, c Underdamping: (results in complex roots) y = e Ax (c 1 cos(Bx) + c 2 sin(Bx))

Section 2.6 Forced Oscillations & Resonance

mx″ + cx′ + kx = F(t) where F(t) = F o cos(ωt) or F(t) = F o sin(ωt)

Undamped Forced Oscillations mx″ + kx = F(t) Let's examine mx″ + kx = F o cos(ωt)

Undamped Forced Oscillations mx″ + kx = F(t) Let's examine mx″ + kx = F o cos(ωt) x c = c 1 cos(ω o t) + c 2 sin(ω o t) (recall ) If ω ≠ ω o then we shall guess that x p = A cos(ωt).

Undamped Forced Oscillations mx″ + kx = F(t) Let's examine mx″ + kx = F o cos(ωt) x c = c 1 cos(ω o t) + c 2 sin(ω o t) (recall ) If ω ≠ ω o then we shall guess that x p = A cos(ωt). Using our guess of x p = A cos(ωt) in the diff. eq. mx″ + kx = F o cos(ωt) we find that

Undamped Forced Oscillations mx″ + kx = F(t) Let's examine mx″ + kx = F o cos(ωt) x c = c 1 cos(ω o t) + c 2 sin(ω o t) (recall ) If ω ≠ ω o then we shall guess that x p = A cos(ωt). Using our guess of x p = A cos(ωt) in the diff. eq. mx″ + kx = F o cos(ωt) we find that So,

Undamped Forced Oscillations mx″ + kx = F(t) Let's examine mx″ + kx = F o cos(ωt) x c = c 1 cos(ω o t) + c 2 sin(ω o t) (recall ) If ω ≠ ω o then we shall guess that x p = A cos(ωt). Using our guess of x p = A cos(ωt) in the diff. eq. mx″ + kx = F o cos(ωt) we find that So, Thus, the general solution is

Undamped Forced Oscillations mx″ + kx = F(t) Let's examine mx″ + kx = F o cos(ωt) x c = c 1 cos(ω o t) + c 2 sin(ω o t) (recall ) If ω ≠ ω o then we shall guess that x p = A cos(ωt). Question - What happens when ω = ω o ? Answer - Resonance

Undamped Forced Oscillations mx″ + kx = F(t) Let's examine mx″ + kx = F o cos(ωt) x c = c 1 cos(ω o t) + c 2 sin(ω o t) (recall ) Suppose ω = ω o. Our guess for x p would be: x p = At cos(ω o t) + Bt sin(ω o t).

Undamped Forced Oscillations mx″ + kx = F(t) Let's examine mx″ + kx = F o cos(ωt) x c = c 1 cos(ω o t) + c 2 sin(ω o t) (recall ) Suppose ω = ω o. Our guess for x p would be: x p = At cos(ω o t) + Bt sin(ω o t). : A = 0

Undamped Forced Oscillations mx″ + kx = F(t) Let's examine mx″ + kx = F o cos(ωt) x c = c 1 cos(ω o t) + c 2 sin(ω o t) (recall ) Suppose ω = ω o. Our guess for x p would be: x p = At cos(ω o t) + Bt sin(ω o t). : A = 0 So,

Undamped Forced Oscillations mx″ + kx = F(t) Let's examine mx″ + kx = F o cos(ωt) x c = c 1 cos(ω o t) + c 2 sin(ω o t) (recall ) Suppose ω = ω o. Our guess for x p would be: x p = At cos(ω o t) + Bt sin(ω o t). : A = 0 So, Thus, the general solution is