This tutorial in the series on mechanical vibrations
is referred to as the ** maths tutorial**.

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 the general method for these
solutions assuming a reasonable
knowledge of college level algebra and basic calculus.

**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

In addition all terms in the equations considered have constant coefficients.

We use** y** to denote the dependent variable and**
x** the independent variable. Thus a solution
means finding a function **y = f(x)** that satisfies
the equation in question.

(In the vibration tutorials the dependent and independent variables are usually x (displacement) and t (time) respectively.)

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

In many cases the starting point for solutions to differential equations 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: * a*r^{2}
+ *b*r + *c* = 0
known as the **characteristic
equation **for the original differential equation.

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

A **particular solution** of the original equation
requires valid values of constants C1 and C2. 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 (C1 and C2).

At this point we must consider the 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.

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.

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

Rearranging exponents of *e *in the RHS of this equation
and then using Euler's formula:

leads to a modified general solution to equation (1) above as follows:

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

Using the identity A.cos(θ) + B.sin(θ) = R.cos(θ - φ)
where R = √(A^{2} + B^{2}) and φ = tan^{-1}
(B /A) expresses equation (4) as:

y = R.e^{αx} cos(βx - φ) where
R = √(C_{1}^{2} + C_{2}^{2})
and φ = tan^{-1} (C_{2 }/ C_{1})
--------- (5)

Equation (5) will be found useful since it expresses y in terms only of cosine.

If (b^{2} - 4ac) = 0
then the characteristic equation has **repeated roots, r**,
but y = C*e*^{rx} is not a complete
general solution to equation (1) because two initial
conditions, and thus two constants, are required. However, it
can be shown that the following expression (known as **reduction of
order**) does provide a complete general solution for repeated roots:

y = C_{1}e^{rx} + C_{2}xe^{rx }
-------- (5)

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

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:

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.

Two trig functions whose first and second derivatives are one or other of these functions are an obvious guess. We propose the following solution:

If the RHS of equation (6) was an exponential
function e^{ax} a "guess" would be y_{p }= Ae^{ax}
and for a polynomial on the RHS a "guess" would be y_{p} =
Ax^{2} +Bx + C

What follows is called the **method of undetermined
coefficients** where we find coefficients A and B by
substituting y_{p}, y_{p}^{/} and y_{p}^{//
}in equation (6).

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)

which gives ( - 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

The general solution y_{g} to the non-homogeneous
equation (6) is a combination of equation (7), the solution y_{c}
to the complementary homogeneous equation, and equation (8), the
particular solution y_{p} to equation (6) which gives:

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