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

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

(\(a_0\), \(a_1\), \(\ldots\), \(a_{n-1}\), \(a_{n}\) 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":

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

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

\[y(x)=c_1e^{r_1x}+c_2e^{r_2x}+\cdots+c_ne^{r_nx} \qquad(0.3)\]

where \(c_1\), \(c_2\), \(\cdots\), \(c_n\) 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), \(r_1\), \(r_2\), \(\cdots\), \(r_n\) are the \(n\) distinct roots of equation \((0.2)\).

And if some of the roots are repeated, for example, say that one of the roots \(r\) is repeated \(k\) times , then the term in the general solution associated with that repeated root would be in the form

\[(c_1+c_2x+c_3x^2+\cdots+c_kx^{k-1})e^{rx} \qquad(0.4)\]

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

\[a_2\dfrac{d^2y}{dx^2}+a_1\dfrac{dy}{dx}+a_0y=0\quad(a_2\neq0)\qquad(1.1)\]

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

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

\[a_2r^2+a_1r+a_0=0\qquad(1.2)\]

This is just a simple quadratic equation, let \(r_1\), \(r_2\) be its two roots, then using Vieta's formula, we have \[r_1+r_2=-\frac{a_1}{a_2} \\ r_1r_2=\frac{a_0}{a_2}\]

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

\[\dfrac{d^2y}{dx^2}-(r_1+r_2)\dfrac{dy}{dx}+r_1r_2y=0\]

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

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

Now, here's the interesting part, notice that if we let \(y_1=\dfrac{dy}{dx}-r_2y\), then \(\dfrac{dy_1}{dx}=\dfrac{d^2y}{dx^2}-r_2\dfrac{dy}{dx}\), the equation above becomes \[\dfrac{dy_1}{dx}-r_1y_1=0\implies\dfrac{dy_1}{dx}=r_1y_1\]

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

\[\frac{1}{y_1}dy_1=r_1dx\]

Integrating both sides:

\[\int\frac{1}{y_1}\,dy_1=\int r_1\,dx\implies \ln |y_1|=r_1x+C_0\]

\[|y_1|=e^{r_1x+C_0}\]

\[y_1=\pm e^{C_0} e^{r_1x}\]

\(\pm e^{C_0}\) is an arbitrary constant, so we'll replace it with \(C_1\):

\[\therefore y_1=C_1 e^{r_1x}\]

Back substituting \(y_1=\dfrac{dy}{dx}-r_2y\): \[\dfrac{dy}{dx}-r_2y=C_1e^{r_1x}\]

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

\[e^{-r_2x}\dfrac{dy}{dx}-r_2e^{-r_2x}y=C_1e^{(r_1-r_2)x}\]

\[\dfrac{d}{dx}\left(e^{-r_2x}y\right)=C_1e^{(r_1-r_2)x}\]

\[\therefore e^{-r_2x}y=\int C_1e^{(r_1-r_2)x}\, dx\qquad(1.3)\]

Case 1: Equation \((1.2)\) has two distinct roots, i.e. \(r_1\neq r_2\)

From \((1.3)\), \[e^{-r_2x}y=\frac{C_1}{r_1-r_2}e^{(r_1-r_2)x}+C_2\] \[y=\frac{C_1}{r_1-r_2}e^{r_1x}+C_2e^{r_2x}\]

Relabeling the constants, we have \[y=c_1e^{r_1x}+c_2e^{r_2x}\]

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

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

Case 2: Equation \((1.2)\) has two equal real roots, i.e. \(r_1=r_2\)

From \((1.3)\), \[e^{-r_1x}y=\int C_1\,dx=C_1x+C_2\]

\[y=C_1xe^{r_1x}+C_2e^{r_1x}\]

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

Third Order

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

\[a_3\dfrac{d^3y}{dx^3}+a_2\dfrac{d^2y}{dx^2}+a_1\dfrac{dy}{dx}+a_0y=0\quad(a_3\neq0)\qquad(2.1)\]

Again, we shall start from its characteristic equation:

\[a_3r^3+a_2r^2+a_1r+a_0=0\qquad(2.2)\]

Ah, a good-ol' cubic equation, let \(r_1\), \(r_2\), \(r_3\) be its three roots, using Vieta's formula, we have \[r_1+r_2+r_3=-\frac{a_2}{a_3} \\ r_1r_2+r_2r_3+r_1r_3=\frac{a_1}{a_3} \\ r_1r_2r_3=-\frac{a_0}{a_3}\]

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

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

\[\dfrac{d^3y}{dx^3}-(r_1+r_2+r_3)\dfrac{d^2y}{dx^2}+(r_1r_2+r_2r_3+r_1r_3)\dfrac{dy}{dx}-r_1r_2r_3y=0\]

\[\dfrac{d^3y}{dx^3}-r_3\dfrac{d^2y}{dx^2}-(r_1+r_2)\dfrac{d^2y}{dx^2}+r_3(r_1+r_2)\frac{dy}{dx}+r_1r_2\dfrac{dy}{dx}-r_1r_2r_3y=0\]

\[\left(\dfrac{d^3y}{dx^3}-r_3\dfrac{d^2y}{dx^2}\right)-(r_1+r_2)\left(\dfrac{d^2y}{dx^2}-r_3\frac{dy}{dx}\right)+r_1r_2\left(\dfrac{dy}{dx}-r_3y\right)=0\]

Substitute \(y_1=\dfrac{dy}{dx}-r_3y\), we get \[\dfrac{d^2y_1}{dx^2}-(r_1+r_2)\dfrac{dy_1}{dx}+r_1r_2y_1=0\qquad(2.3)\]

Déjà vu? Hey! That's because we're back to the second order case again! In fact, if we let \(r_1+r_2=-\frac{a'_1}{a'_2}\) and \(r_1r_2=\frac{a'_0}{a'_2}\) (\(a'_0\), \(a'_1\), \(a'_2\) are constants, \(a'_2\neq0\)), we have \[\dfrac{d^2y_1}{dx^2}+\frac{a'_1}{a'_2}\cdot\dfrac{dy_1}{dx}+\frac{a'_0}{a'_2}\cdot y=0 \implies a'_2\dfrac{d^2y_1}{dx^2}+a'_1\dfrac{dy_1}{dx}+a'_0y=0\] with \(r_1\), \(r_2\) being the roots of its characteristic equation \(a'_2r^2+a'_1r+a'_0=0\).

Hence we're dealing with a second order homogeneous linear DE with constant coefficients, which we know its general solution, and because \(r_1\neq r_2\), \[y_1=\dfrac{dy}{dx}-r_3y=C_1e^{r_1x}+C_2e^{r_2x}\]

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

Integrating both sides and multiplying both sides by \(e^{r_3x}\), we have \[y=\frac{C_1}{r_1-r_3}e^{r_1x}+\frac{C_2}{r_2-r_3}e^{r_2x}+C_3e^{r_3x}\]

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

Case 2: Equation \((2.2)\) has one repeated root, without loss of generality, let's assume that \(r_1=r_3\)

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

Case 3: Equation \((2.2)\) has 3 equal real roots, i.e. \(r_1=r_2=r_3\)

Similar to case 1, but when solving for \(y_1\) in \((2.3)\) would instead give \[y_1=\dfrac{dy}{dx}-r_1y=(C_1+C_2x)e^{r_1x}\]

Multiplying both sides by its integrating factor \(e^{\int -r_1\, dx}=e^{-r_1x}\) gives \[\dfrac{d}{dx}(e^{-r_1x}y)=C_1+C_2x\]

Integrating both sides: \[e^{-r_1x}y=C_1x+\frac{C_2}{2}x^2+C_3\]

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

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 \((0.3)\), a linear sequence of powers of \(e\), which makes sense, since \(e\) 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 \((0.1)\) might be in the form \(e^{rx}\).

Things also get a bit more interesting when there are repeated roots. You might have noticed in \((1.3)\) and \((2.4)\), when there are repeated roots, they cause the exponent of the terms \(Ce^{ax}\) to become zero, causing it to be equal to some constant \(C\). When constants are integrated, linear terms of \(x\) appear, in higher orders, the \(x\) terms are integrated several more times and become \(x^2\) terms, then \(x^3\), \(x^4\), \(\ldots\). This is one of the reasons integer powers of \(x\) appear in \((0.4)\) 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
2 weeks, 6 days ago

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@Kenneth Tan Thanks for this!!!

Aaghaz Mahajan - 2 weeks, 1 day ago

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Glad it helped :)

Kenneth Tan - 2 weeks, 1 day ago

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Do computer science students take calculus? I wonder.

Donglin Loo - 2 weeks, 1 day ago

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Usually they take calculus but is not too in depth.

Kenneth Tan - 2 weeks, 1 day 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 week, 2 days ago

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