This series of tutorials on the subject of mechanical vibrations hinges on solutions to a particular type of differential equation defined as **ordinary, second order, linear, homogeneous or non-homogeneous**. The purpose of this tutorial is to explain some general methods for finding these solutions. A reasonable knowledge of college level algebra and basic calculus is assumed.

**ordinary**means that all derivatives of the one dependent variable in the equation are with respect to the same independent variable (i.e. there are no partial derivatives)**second order**means that the highest derivative in the equation is of the second order**linear**means that the exponent of each term containing the dependent variable in the equation is one**homogeneous**equations of this type equal zero;**non-homgeneous**equations do not

Additionally, all terms in the equations considered have **constant coefficients**.

In this tutorial **y** denotes dependent variables and **x** independent variables. Thus a solution means finding a function **y = f(x)** that satisfies the equation in question. In subsequent tutorials the dependent and independent variables are typically **x** (displacement) and **t** (time) respectively.

### Solutions to homogeneous equations

We first consider a solution to the homogeneous equation which has the general form:

In many cases the starting point for a solution to a differential equation is to find a function that could be the basis for a solution.

Replacing y and its derivatives in (1) above with expressions in e^{rx} gives:

Because e^{rx} cannot equal zero this equation reduces to: **ar ^{2} + br + c = 0** known as the

**characteristic equation**of the original differential equation.

It can be shown that any linear combination of these solutions is also a solution. Hence the **general solution** can be expressed as:

A **particular solution** of the original equation requires values for constants C_{1} and C_{2}. These are found by setting **initial conditions** for y and dy/dx when x = 0 and substituting these values in equations (2a) and (2b) giving two equations with two unknowns (C_{1} and C_{2}).

We now examine solutions r_{1} and r_{2} of the characteristic equation taking account of the value of the term (b^{2} - 4ac) in the quadratic formula.

**1. (b ^{2} - 4ac) > 0**

If (b^{2} - 4ac) > 0 then r_{1} and r_{2} are real numbers and the general solution can be obtained directly from equation (2) above.

**2. (b ^{2} - 4ac) < 0 **

If (b^{2} - 4ac) < 0 then r_{1} and r_{2} can be expressed as a pair of **complex conjugate** numbers as follows:

We now expand equation (3) to give:

Now replace the constant terms in the above equation with "new" constants C_{1} and C_{2} respectively.

Equation(4) has no terms in ** i** making it much easier to apply as a general solution than equation (3) above

Further simplification expresses the general solution of equation (4) as a single cosine function in the following way.

From the identity **A.cos(θ) + B.sin(θ) = R.cos(θ - φ**)

where **R = √(A ^{2} + B^{2})** and

**φ = tan**

^{-1}(B /A)it follows that:

**y = R.e ^{αx} cos(βx - φ) --------- (5)**

where **R = √(C _{1}^{2} + C_{2}^{2})** and

**φ = tan**

^{-1}(C_{2}/ C_{1})**3. (b ^{2} - 4ac) = 0 **

If **(b ^{2} - 4ac) = 0** then the characteristic equation has

**repeated roots, r**and the general solution of equation 2(a):

becomes **y = C.e ^{rx} ------ (2c)** with constant

**C**.

This is not a complete general solution because two initial conditions, and thus two constants, are required. However, it can be shown that equation (5) below, known as **reduction of order**, provides a template for solutions where there are repeated roots:

**y = C _{1}e^{rx} + C_{2}x.e^{rx} -------- (5)**

In subsequent tutorials we use the general solutions above to solve the homogeneous equations of motion for **free vibrating systems**.

### Solutions to non-homogeneous equations

The **general solution** to this form of equation is:

**the general solution for the complementary homogeneous equation**

**+**

**+**

**the particular solution for the non-homogeneous equation**

This is understood better by working through an example for the following non-homogeneous equation (6):

*We use a cosine function f(x) = 2cos(3x) in our example because a function of this type arises in non-homogeneous equations of motion for forced vibrating systems.*

**Step 1 - find the general solution to the complementary homogeneous equation**

**Step 2 - find the particular solution to the non-homogeneous equation **

A starting point is to find two functions whose first and second derivatives replicate each other. Sine and cosine functions give a possible particular solution to equation (6) as follows:

What follows is called the **method of undetermined coefficients** where we find coefficients A and B by substituting **y _{p}, y_{p}^{/}** and

**y**in equation (6).

_{p}^{//}**y _{p}^{/} = - 3.A.sin(3x) + 3.B.cos(3x)**

**y _{p}^{//} = -9.A.cos(3x) - 9.B.sin(3x)**

Substituting in equation (6) gives:

**-9.A.cos(3x) - 9.B.sin(3x) -6.A.sin(3x) +6.B.cos(3x) -3. A.cos(3x) -3.B.sin(3x) = 2.cos(3x)**

giving **(- 12A + 6.B).cos(3x) + ( - 6.A - 12B).sin(3x) = 2.cos(3x)**

equating coefficients for cos(3x) and sin(3x) on both sides of the equation gives:

**(- 12A + 6.B) = 2 and (- 6.A - 12B) = 0**

**Step 3 - combine solutions **

The general solution **y _{g}** to the non-homogeneous equation (6) is a combination of the solution

**y**to the complementary homogeneous equation and the particular solution

_{c}**y**to equation (6) which gives:

_{p}Next: Free vibrations

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