Derivation of D.E. solution

i'm writing this post mostly to show how we will arrive at the solution to the differential equation (D.E.) that we will encounter for SHM (Simple Harmonic Motion) in my next post. this is so that nothing is left out of the proof and my reader(s?) will not feel lost at all! with this proof and the following post, you should have everything you need to solve the SHM differential equation.

but this is not just used to solve the SHM equation! do note that this solution can be applied to many differential equations, and really rather easily! maybe you'll come across one and this post will have helped you solve it! (well, i can only dream).

there probably will not be as much step-by-step explanation in this post, mostly just math. but if you follow carefully, and i don't make any mistakes, the math should speak for itself.
so here goes:

first we need to mention that this is a homogeneous linear D.E. with constant coefficients which can have any order derivative.
i will present mine as a D.E. with a 2nd order derivative because that is the most basic and relevant example, and it can easily be expanded to any higher derivatives with little (or no) extra work.

consider, then the D.E.:
d²x/dt² + 2a* dx/dt + bx = 0, or
x'' + 2ax' + bx = 0, where a and b are constants. this can be written in operator notation as:
(D² + 2aD + b) x = 0
and if my convention for operator notation is different that your, D is the differential operator, d/dt. and so dx/dt is written Dx or D(x).

now, in order to proceed, we need to know something about linear D.E.s with constant coefficients - which is where Euler will come in and help us. something known as an 'auxiliary equation' is used to classify this D.E. and quickly arrive at its solution. the aux. equ. is simply:
r² + 2ar + b = 0
regardless of the particular solutions, this quadratic has 2 roots, r₁ and r₂, such that
(r - r₁) (r - r₂) = 0
(and it may have more for a higher order differential)

it follows then that our D.E. can be written as:
(D - r₁) (D - r₂) x = 0

now, let u = (D - r₂) x, and we can write our equation as:
(D - r₁) u = 0, and solve for u
du/dt - r₁u = 0
du/dt = r₁u
du/u = r₁dt, and integrating...
∫du/u = ∫ r₁dt + C₁''
ln|u| = r₁t + C₁''
|u| = e^( r₁t + C₁''), and let |C₁'| = e^C₁''
u = C₁' e^( r₁t)

then, since:
(D - r₂) x = u
dx/dt - r₂x = C₁' e^(r₁t), or
dx/dt + (- r₂) x = C₁' e^(r₁t), which is of a form which has a known solution. but i'm not going to even cut any corners there. i'll walk through it.

let p(t) = -r₂, q(t) = C₁' e^(r₁t), such that our equation is:
x' + px = q
now let μ = e^[∫p(t)dt] = e^[∫-r₂dt] = e^(-r₂t)
note also that dμ/dt = -r₂ e^(-r₂t) = μp
now multiply through our equation with μ (which is okay since μ ≠ 0 ∀t):
μx' + μpx = μq, or
μ (dx/dt) + x (dμ/dt) = μq, which should look like the product rule for differentiation to you, thus:
d(μx)/dt = μq, or
d(μx) = μqdt, and integrating:
∫d(μx) = ∫μqdt + C₂
μx = ∫μqdt + C₂, and substituting back in:
e^(-r₂t) x = ∫ e^(-r₂t) C₁' e^(r₁t) dt + C₂
e^(-r₂t) x = C₁' ∫ e^[(r₁ - r₂) t] dt + C₂

here, we will observe 3 special cases for evaluating the integral ∫ e^[(r₁ - r₂) t] dt:
CASE 1: r₁ = r₂ = r
e^(-rt) x = C₁' ∫ e^[0 t] dt + C₂
e^(-rt) x = C₁'t + C₂, (and let C₁ = C₁')
x(t) = e^(rt) (C₁t + C₂)

CASE 2: r₁ ≠ r₂
e^(-r₂t) x = C₁' ∫ e^[(r₁ - r₂) t] dt + C₂
e^(-r₂t) x = C₁'/(r₁ - r₂) * e^[(r₁ - r₂) t] + C₂, (and let C₁ = C₁'/(r₁ - r₂))
e^(-r₂t) x = C₁ e^[(r₁ - r₂) t] + C₂
x(t) = C₁ e^(r₁t) + C₂ e^(r₂t)

CASE 3: r = α ± iβ (the case for complex roots of the aux. equ.)
let r₁ = α + iβ, r₂ = α - iβ
e^[-(α - iβ) t] x = C₁' ∫ e^[(2iβ) t] dt + C₂
e^[-(α - iβ) t] x = C₁'/2iβ * e^[(2iβ) t] + C₂, (and let C₁ = C₁'/2iβ)
e^[-(α - iβ) t] x = C₁ e^[(2iβ) t] + C₂
x = e^[(α - iβ) t] * C₁ e^[(2iβ) t] + e^[(α - iβ) t] * C₂
x(t) = e^(αt) [ C₁e^(iβt) + C₂e^(-iβt) ]
(to make things a little more interesting still, convert this to sin and cos using Euler's formula! i'll have to do this in the next post anyway)
(hint: just call (C₁ + C₂) and (C₁ - C₂), C₁* and C₂*, respectively)

and there we go! 3 comprehensive solutions to ANY homogeneous linear D.E. with constant coefficients. but what if you have higher order derivatives and more roots in your aux. equ., you say? well, those can simply be handled like so:
CASE 1 will become:
x(t) = e^(rt) (C₁t² + C₂t + C₃), and so forth for more roots...
CASE 2 will (easily) become:
x(t) = C₁ e^(r₁t) + C₂ e^(r₂t) + C₃ e^(r₃t) ... for more roots
and CASE 3 you don't really have to worry about because you will never have complex roots that don't come in pairs (unless your coefficients are complex)

what'll really get you is more multiplicities in your roots! consider the equation:
(D² + 3D + 2)² x = 0, which you can write:
(D + 2)² (D + 1)² x = 0. the solutions for this turn out to be a combination of CASE 1 and CASE 2:
x(t) = e^(-2t) (C₄t + C₃) + e^(-t) (C₂t + C₁)
verify yourself!
the cool part is that, with practice, you should be able to solve these differential equations in 5 or 10 seconds in your head! amaze your friends! trust me, it's very fun :] all you need to know how to do is factor polynomials!

what about the following?:
(D² + 4D + 5)² (D² - 5D + 4)³ x = 0
well, it may take you more than 5 or 10 seconds to actually write it, but you should have no problem coming up with an answer very quickly!

also, something else to note as we go through these examples: always make sure the number of constants matches up with the order of differentials in your D.E.! if there is a 2nd order derivative in your D.E., you should have a C₁ and a C₂. this is, of course, easily understandable since you would normally have to do 6 integrations for a D.E. with a 6th order derivative, and thus giving you 6 constants of integration.

so there, i hope that this will completely resolve the question of how to solve the SHM equation and even some general solutions for a plethora of differential equations! enjoy! (challenge: solve the differential equation which governs a dampened oscillation!)