Semaj Christian

Answered

2022-06-27

One thing I've hated about differential equations is how I need to guess the form of the solution.

e.g. it's easy to show that solutions to a linear constant-coefficient differential equation such as

${y}^{\u2034}+2{y}^{\u2033}+3{y}^{\prime}+4y=0$

have the form of (some linear combination of) exponentials:

You can just plug in $y={e}^{ax+b}$ and show that the equation is satisfiable.

But I feel there is something wrong if I must find the solutions through guess-and-check.

Yet what I was never told, and what I seem to never be able to find from looking online, is how to derive this fact rigorously, when I don't already have the intuition necessary to guess the form of the solution?

Sitting down and working through some math, I've come up with some nonsense that works quite beautifully:

1. Place the homogeneous equation into the following matrix form:

$\overrightarrow{y}{\phantom{\rule{thinmathspace}{0ex}}}^{\prime}=A\overrightarrow{y}$

where, for example, we have $\overrightarrow{y}={\left[\begin{array}{ccc}y& {y}^{\prime}& \dots \end{array}\right]}^{T}$

2. Drop the arrows and pretend everything is a scalar:

${y}^{\prime}=Ay$

3. Separate the, uh, variables:

$\frac{dy}{dx}=Ay$

$\frac{dy}{y}=A\phantom{\rule{thinmathspace}{0ex}}dx$

4. Integrate:

$\int \frac{dy}{y}=\int A\phantom{\rule{thinmathspace}{0ex}}dx$

$\mathrm{ln}y=Ax+b$

$y={e}^{Ax+b}$

5. Profit!

How do I derive it properly?

e.g. it's easy to show that solutions to a linear constant-coefficient differential equation such as

${y}^{\u2034}+2{y}^{\u2033}+3{y}^{\prime}+4y=0$

have the form of (some linear combination of) exponentials:

You can just plug in $y={e}^{ax+b}$ and show that the equation is satisfiable.

But I feel there is something wrong if I must find the solutions through guess-and-check.

Yet what I was never told, and what I seem to never be able to find from looking online, is how to derive this fact rigorously, when I don't already have the intuition necessary to guess the form of the solution?

Sitting down and working through some math, I've come up with some nonsense that works quite beautifully:

1. Place the homogeneous equation into the following matrix form:

$\overrightarrow{y}{\phantom{\rule{thinmathspace}{0ex}}}^{\prime}=A\overrightarrow{y}$

where, for example, we have $\overrightarrow{y}={\left[\begin{array}{ccc}y& {y}^{\prime}& \dots \end{array}\right]}^{T}$

2. Drop the arrows and pretend everything is a scalar:

${y}^{\prime}=Ay$

3. Separate the, uh, variables:

$\frac{dy}{dx}=Ay$

$\frac{dy}{y}=A\phantom{\rule{thinmathspace}{0ex}}dx$

4. Integrate:

$\int \frac{dy}{y}=\int A\phantom{\rule{thinmathspace}{0ex}}dx$

$\mathrm{ln}y=Ax+b$

$y={e}^{Ax+b}$

5. Profit!

How do I derive it properly?

Answer & Explanation

scipionhi

Expert

2022-06-28Added 25 answers

So, most of the math you've seen up to now is very simple, by mathematical standards. The sorts of problems that come up in algebra, trig, and calculus courses can often be solved just by starting from an equation, applying a set of transformations to both sides of the equation, and arriving at an answer.

Most of math doesn't work that way at all. I mean, consider Euclid's proof that there are infinitely many prime numbers, for instance. Would you ask something like

"Why did I have to guess that looking at ${p}_{1}{p}_{2}\cdots {p}_{n}+1$ and its factorization was the right thing to do? Isn't there some way to just derive this automatically?"

No, of course not. There's nothing nonrigorous about Euclid's proof, and there's nothing nonrigorous about observing that the $k$-th derivative of ${e}^{ax+b}$ is ${a}^{k}{e}^{ax+b}$, and that taking that together with the fundamental theorem of algebra will give us all the solutions we seek.

Most of math doesn't work that way at all. I mean, consider Euclid's proof that there are infinitely many prime numbers, for instance. Would you ask something like

"Why did I have to guess that looking at ${p}_{1}{p}_{2}\cdots {p}_{n}+1$ and its factorization was the right thing to do? Isn't there some way to just derive this automatically?"

No, of course not. There's nothing nonrigorous about Euclid's proof, and there's nothing nonrigorous about observing that the $k$-th derivative of ${e}^{ax+b}$ is ${a}^{k}{e}^{ax+b}$, and that taking that together with the fundamental theorem of algebra will give us all the solutions we seek.

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