More On Homogeneous Linear Differential Equations With Constant Coefficients

OH BOY who's ready to board the hype train for differential equations!?

*cricket sounds*

*more cricket sounds*

*awkward silence*

Anyway... in case you're wondering, what we're gonna do today is we're gonna look into homogeneous linear differential equations with constant coefficients, that is, differential equations in the form

andnydxn+an1dn1ydxn1++a1dydx+a0y=0(0.1)a_n\dfrac{d^ny}{dx^n}+a_{n-1}\dfrac{d^{n-1}y}{dx^{n-1}}+\cdots+a_1\dfrac{dy}{dx}+a_0y=0\qquad(0.1)

(a0a_0, a1a_1, \ldots, an1a_{n-1}, ana_{n} are arbitrary real constants), and we're gonna talk about their general solutions! (yay)

By the way, if you don't know about homogeneous linear differential equations (really long name I know) and how to solve them, look it up here.

Now, there is this thing with a nice name called "characteristic equation" or "auxiliary equation":

anrn+an1rn1++a1r+a0=0(0.2)a_nr^n+a_ {n-1}r^{n-1}+\cdots+a_1r+a_0=0\qquad(0.2) which determines the nature of the general solution of (0.1)(0.1).

In fact, we have this theorem stating that when all roots of equation (0.2)(0.2) are distinct, then a general solution for equation (0.1)(0.1) will be in the form

y(x)=c1er1x+c2er2x++cnernx(0.3)y(x)=c_1e^{r_1x}+c_2e^{r_2x}+\cdots+c_ne^{r_nx} \qquad(0.3)

where c1c_1, c2c_2, \cdots, cnc_n are arbitrary constants (may be real or complex, if there are complex roots, they are also rewritten to the trigonometric form using Euler's formula but that's beside the main point), r1r_1, r2r_2, \cdots, rnr_n are the nn distinct roots of equation (0.2)(0.2).

And if some of the roots are repeated, for example, say that one of the roots rr is repeated kk times , then the term in the general solution associated with that repeated root would be in the form

(c1+c2x+c3x2++ckxk1)erx(0.4)(c_1+c_2x+c_3x^2+\cdots+c_kx^{k-1})e^{rx} \qquad(0.4)

Now two questions might arise from our heads, why exactly must it be in the form (0.3)(0.3)? And where did the (c1+c2x+c3x2++ckxk1)(c_1+c_2x+c_3x^2+\cdots+c_kx^{k-1}) part in (0.4)(0.4) come from?

There's one explanation that uses the principle of superposition (not the one involving waves!), but here I'm gonna present you an alternative approach to derive (0.3)(0.3) and (0.4)(0.4), so hold on tight and wear your seatbelts!

Second Order

Before we do anything crazy, let's look at something simpler, a second order homogeneous linear DE with constant coefficients:

a2d2ydx2+a1dydx+a0y=0(a20)(1.1)a_2\dfrac{d^2y}{dx^2}+a_1\dfrac{dy}{dx}+a_0y=0\quad(a_2\neq0)\qquad(1.1)

We will now try to derive a general solution that satisfies (1.1)(1.1).

Instead of "guessing" that the solution would look like the form erxe^{rx}, let's instead start at its characteristic equation shall we?

a2r2+a1r+a0=0(1.2)a_2r^2+a_1r+a_0=0\qquad(1.2)

This is just a simple quadratic equation, let r1r_1, r2r_2 be its two roots, then using Vieta's formula, we have r1+r2=a1a2r1r2=a0a2r_1+r_2=-\frac{a_1}{a_2} \\ r_1r_2=\frac{a_0}{a_2}

Back to (1.1)(1.1), since a20a_2\neq0, we have d2ydx2+a1a2dydx+a0a2y=0\dfrac{d^2y}{dx^2}+\frac{a_1}{a_2}\cdot\dfrac{dy}{dx}+\frac{a_0}{a_2}\cdot y=0

d2ydx2(r1+r2)dydx+r1r2y=0\dfrac{d^2y}{dx^2}-(r_1+r_2)\dfrac{dy}{dx}+r_1r_2y=0

d2ydx2r2dydxr1dydx+r1r2y=0\dfrac{d^2y}{dx^2}-r_2\dfrac{dy}{dx}-r_1\dfrac{dy}{dx}+r_1r_2y=0

(d2ydx2r2dydx)r1(dydxr2y)=0\left(\dfrac{d^2y}{dx^2}-r_2\dfrac{dy}{dx}\right)-r_1\left(\dfrac{dy}{dx}-r_2y\right)=0

Now, here's the interesting part, notice that if we let y1=dydxr2yy_1=\dfrac{dy}{dx}-r_2y, then dy1dx=d2ydx2r2dydx\dfrac{dy_1}{dx}=\dfrac{d^2y}{dx^2}-r_2\dfrac{dy}{dx}, the equation above becomes dy1dxr1y1=0    dy1dx=r1y1\dfrac{dy_1}{dx}-r_1y_1=0\implies\dfrac{dy_1}{dx}=r_1y_1

Would you look at that! What we have now is a separable differential equation, which can be easily solved:

1y1dy1=r1dx\frac{1}{y_1}dy_1=r_1dx

Integrating both sides:

1y1dy1=r1dx    lny1=r1x+C0\int\frac{1}{y_1}\,dy_1=\int r_1\,dx\implies \ln |y_1|=r_1x+C_0

y1=er1x+C0|y_1|=e^{r_1x+C_0}

y1=±eC0er1xy_1=\pm e^{C_0} e^{r_1x}

±eC0\pm e^{C_0} is an arbitrary constant, so we'll replace it with C1C_1:

y1=C1er1x\therefore y_1=C_1 e^{r_1x}

Back substituting y1=dydxr2yy_1=\dfrac{dy}{dx}-r_2y: dydxr2y=C1er1x\dfrac{dy}{dx}-r_2y=C_1e^{r_1x}

Brilliant! We now have a first order linear differential equation, to solve this, we multiply both sides by its integrating factor er2dx=er2xe^{\int -r_2\, dx}=e^{-r_2x}:

er2xdydxr2er2xy=C1e(r1r2)xe^{-r_2x}\dfrac{dy}{dx}-r_2e^{-r_2x}y=C_1e^{(r_1-r_2)x}

ddx(er2xy)=C1e(r1r2)x\dfrac{d}{dx}\left(e^{-r_2x}y\right)=C_1e^{(r_1-r_2)x}

er2xy=C1e(r1r2)xdx(1.3)\therefore e^{-r_2x}y=\int C_1e^{(r_1-r_2)x}\, dx\qquad(1.3)

Case 1: Equation (1.2)(1.2) has two distinct roots, i.e. r1r2r_1\neq r_2

From (1.3)(1.3), er2xy=C1r1r2e(r1r2)x+C2e^{-r_2x}y=\frac{C_1}{r_1-r_2}e^{(r_1-r_2)x}+C_2 y=C1r1r2er1x+C2er2xy=\frac{C_1}{r_1-r_2}e^{r_1x}+C_2e^{r_2x}

Relabeling the constants, we have y=c1er1x+c2er2xy=c_1e^{r_1x}+c_2e^{r_2x}

which is exactly what we wanted, it is a general solution that satisfies (1.1)(1.1), and is in the same form as (0.3)(0.3)

Let's now see what happens if the characteristic equation has two equal real roots.

Case 2: Equation (1.2)(1.2) has two equal real roots, i.e. r1=r2r_1=r_2

From (1.3)(1.3), er1xy=C1dx=C1x+C2e^{-r_1x}y=\int C_1\,dx=C_1x+C_2

y=C1xer1x+C2er1xy=C_1xe^{r_1x}+C_2e^{r_1x}

Relabeling the constants and rearranging, we have y=(c1+c2x)er1xy=(c_1+c_2x)e^{r_1x} which is also what we wanted, it is a general solution and it is in the same form as (0.4)(0.4).

Third Order

Now what about third order homogeneous linear DEs with constant coefficients? Fear not! We're about to find out!

a3d3ydx3+a2d2ydx2+a1dydx+a0y=0(a30)(2.1)a_3\dfrac{d^3y}{dx^3}+a_2\dfrac{d^2y}{dx^2}+a_1\dfrac{dy}{dx}+a_0y=0\quad(a_3\neq0)\qquad(2.1)

Again, we shall start from its characteristic equation:

a3r3+a2r2+a1r+a0=0(2.2)a_3r^3+a_2r^2+a_1r+a_0=0\qquad(2.2)

Ah, a good-ol' cubic equation, let r1r_1, r2r_2, r3r_3 be its three roots, using Vieta's formula, we have r1+r2+r3=a2a3r1r2+r2r3+r1r3=a1a3r1r2r3=a0a3r_1+r_2+r_3=-\frac{a_2}{a_3} \\ r_1r_2+r_2r_3+r_1r_3=\frac{a_1}{a_3} \\ r_1r_2r_3=-\frac{a_0}{a_3}

Case 1: Equation (2.2)(2.2) has 3 distinct roots

As a30a_3\neq0, from (2.1)(2.1), we have d3ydx3+a2a3d2ydx2+a1a3dydx+a0a3y=0\dfrac{d^3y}{dx^3}+\frac{a_2}{a_3}\cdot\dfrac{d^2y}{dx^2}+\frac{a_1}{a_3}\cdot\dfrac{dy}{dx}+\frac{a_0}{a_3}\cdot y=0

d3ydx3(r1+r2+r3)d2ydx2+(r1r2+r2r3+r1r3)dydxr1r2r3y=0\dfrac{d^3y}{dx^3}-(r_1+r_2+r_3)\dfrac{d^2y}{dx^2}+(r_1r_2+r_2r_3+r_1r_3)\dfrac{dy}{dx}-r_1r_2r_3y=0

d3ydx3r3d2ydx2(r1+r2)d2ydx2+r3(r1+r2)dydx+r1r2dydxr1r2r3y=0\dfrac{d^3y}{dx^3}-r_3\dfrac{d^2y}{dx^2}-(r_1+r_2)\dfrac{d^2y}{dx^2}+r_3(r_1+r_2)\frac{dy}{dx}+r_1r_2\dfrac{dy}{dx}-r_1r_2r_3y=0

(d3ydx3r3d2ydx2)(r1+r2)(d2ydx2r3dydx)+r1r2(dydxr3y)=0\left(\dfrac{d^3y}{dx^3}-r_3\dfrac{d^2y}{dx^2}\right)-(r_1+r_2)\left(\dfrac{d^2y}{dx^2}-r_3\frac{dy}{dx}\right)+r_1r_2\left(\dfrac{dy}{dx}-r_3y\right)=0

Substitute y1=dydxr3yy_1=\dfrac{dy}{dx}-r_3y, we get d2y1dx2(r1+r2)dy1dx+r1r2y1=0(2.3)\dfrac{d^2y_1}{dx^2}-(r_1+r_2)\dfrac{dy_1}{dx}+r_1r_2y_1=0\qquad(2.3)

Déjà vu? Hey! That's because we're back to the second order case again! In fact, if we let r1+r2=a1a2r_1+r_2=-\frac{a'_1}{a'_2} and r1r2=a0a2r_1r_2=\frac{a'_0}{a'_2} (a0a'_0, a1a'_1, a2a'_2 are constants, a20a'_2\neq0), we have d2y1dx2+a1a2dy1dx+a0a2y=0    a2d2y1dx2+a1dy1dx+a0y=0\dfrac{d^2y_1}{dx^2}+\frac{a'_1}{a'_2}\cdot\dfrac{dy_1}{dx}+\frac{a'_0}{a'_2}\cdot y=0 \implies a'_2\dfrac{d^2y_1}{dx^2}+a'_1\dfrac{dy_1}{dx}+a'_0y=0 with r1r_1, r2r_2 being the roots of its characteristic equation a2r2+a1r+a0=0a'_2r^2+a'_1r+a'_0=0.

Hence we're dealing with a second order homogeneous linear DE with constant coefficients, which we know its general solution, and because r1r2r_1\neq r_2, y1=dydxr3y=C1er1x+C2er2xy_1=\dfrac{dy}{dx}-r_3y=C_1e^{r_1x}+C_2e^{r_2x}

This is a first order linear differential equation, multiplying both sides by its integrating factor er3dx=er3xe^{\int -r_3\, dx}=e^{-r_3x}, we will get ddx(er3xy)=C1e(r1r3)x+C2e(r2r3)x(2.4)\dfrac{d}{dx}(e^{-r_3x}y)=C_1e^{(r_1-r_3)x}+C_2e^{(r_2-r_3)x}\qquad(2.4)

Integrating both sides and multiplying both sides by er3xe^{r_3x}, we have y=C1r1r3er1x+C2r2r3er2x+C3er3xy=\frac{C_1}{r_1-r_3}e^{r_1x}+\frac{C_2}{r_2-r_3}e^{r_2x}+C_3e^{r_3x}

Relabeling the constants: y=c1er1x+c2er2x+c3er3xy=c_1e^{r_1x}+c_2e^{r_2x}+c_3e^{r_3x} \, _\square

Case 2: Equation (2.2)(2.2) has one repeated root, without loss of generality, let's assume that r1=r3r_1=r_3

In this case, all the steps will be identical to case 1 until up to (2.4)(2.4), where integrating both sides of (2.4)(2.4) would instead give er1xy=C1x+C2r2r3er2x+C3e^{-r_1x}y=C_1x+\frac{C_2}{r_2-r_3}e^{r_2x}+C_3 Multiplying both sides by er1xe^{r_1x} and relabeling the constants, we have y=(c1+c2x)er1x+c3er2xy=(c_1+c_2x)e^{r_1x}+c_3e^{r_2x}\, _\square

Case 3: Equation (2.2)(2.2) has 3 equal real roots, i.e. r1=r2=r3r_1=r_2=r_3

Similar to case 1, but when solving for y1y_1 in (2.3)(2.3) would instead give y1=dydxr1y=(C1+C2x)er1xy_1=\dfrac{dy}{dx}-r_1y=(C_1+C_2x)e^{r_1x}

Multiplying both sides by its integrating factor er1dx=er1xe^{\int -r_1\, dx}=e^{-r_1x} gives ddx(er1xy)=C1+C2x\dfrac{d}{dx}(e^{-r_1x}y)=C_1+C_2x

Integrating both sides: er1xy=C1x+C22x2+C3e^{-r_1x}y=C_1x+\frac{C_2}{2}x^2+C_3

Multiplying both sides by er1xe^{r_1x} and relabeling the constants, we have y=(c1+c2x+c3x2)er1xy=(c_1+c_2x+c_3x^2)e^{r_1x}\, _\square

Some Observations

We have shown that if the roots of a characteristic equation of a second or third order homogeneous linear DE with constant coefficients are all distinct, then its general solution will be in the form (0.3)(0.3), a linear sequence of powers of ee, which makes sense, since ee comes up a lot in calculus, it's a special magic number with special properties. It also makes it reasonable to guess that solutions to (0.1)(0.1) might be in the form erxe^{rx}.

Things also get a bit more interesting when there are repeated roots. You might have noticed in (1.3)(1.3) and (2.4)(2.4), when there are repeated roots, they cause the exponent of the terms CeaxCe^{ax} to become zero, causing it to be equal to some constant CC. When constants are integrated, linear terms of xx appear, in higher orders, the xx terms are integrated several more times and become x2x^2 terms, then x3x^3, x4x^4, \ldots. This is one of the reasons integer powers of xx appear in (0.4)(0.4) when repeated roots are involved.

Higher Orders?

Of course, the derivations above can be extended to higher orders, an inductive way to prove for all orders might also exist, but this note is already too long and is left as an exercise to the reader :D (I'm so nice).

This might be the longest note I've ever written, so if you made it all the way here, then you are awesome. :)

Toodles, gotta go!

Note by Kenneth Tan
1 year, 1 month ago

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@Kenneth Tan Thanks for this!!!

Aaghaz Mahajan - 1 year, 1 month ago

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Glad it helped :)

Kenneth Tan - 1 year, 1 month ago

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Some criticisms:

  1. In the second-order case, where you have two distinct roots, what happens when the two roots are "complex"? If you are looking for solutions in the "real number field", you will need to do some work to adjust it.

  2. When you go to higher-orders, using the method of characteristic equations will fail once you get to degree 5 and higher (by the Abel-Ruffini theorem). Because of this, you may need to adjust your methods. I am not sure how to tackle this though; not sure if using a linear algebraic method of solving it as a system of ODEs would solve your problem, as finding the eigenvalues generally requires solving such a polynomial. In any case, you may think about it.

In any case, a nice primer for anyone looking to understand homogeneous linear ODEs, especially in the third-order case. Ultimately, these methods will fall very short from a pure mathematical perspective, as the calculations become (almost) infinitely more complicated.

A Brilliant Member - 1 year ago

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Do computer science students take calculus? I wonder.

donglin loo - 1 year, 1 month ago

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Usually they take calculus but is not too in depth.

Kenneth Tan - 1 year, 1 month ago

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Fair enough. After all, computers don't mix well with infinitely many tasks. Looking at you integrals.

Chris Rather not say - 1 year, 1 month ago

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