## Solving linear differential equations by "factoring"

I have thought of an interesting way of solving n-th order linear differential equations (with constant coefficients) by imitating the way we solve n-th order polynomials, that is by "factoring" it into a "product" of simple 1st order linear differential equations. Does this method already exist? I haven't been able to prove it rigorously, but the case n=3 and 2 convinced me that it is reasonable to say it's true.

We have the n-th order linear diff. eq.:
$$a_{n}y^{(n)}(x)+a_{n-1}y^{(n-1)}(x)+ ... +a_{0}y(x) = f(x)$$
or in terms of the differential operator: $$(a_nD^n+a_{n-1}D^{n-1}+ ... + a_0)[y(x)] = f(x)$$

My claim is that any such equation can be "factored" into
$$a_n(D-\lambda_n)(D-\lambda_{n-1})...(D-\lambda_1)[y(x)] = f(x)$$

where $D$ is the differential operator, and $\lambda_{i}$ is the i-th root of the polynomial equation $$a_{n}x^{n}+a_{n-1}x^{n-1}+ ... + a_{0} = 0$$

and the "multiplication" is actually the successive composition of the $D-\lambda_i$ operations, which take the derivative of the argument and then substract $\lambda_i$ times the argument.

Then by naming the function resulting from the i-th operation $\phi_i(x)$ we have the equation $$a_n\phi_n(x) = f(x)$$ or $$a_n\phi_{n-1}^'(x) - a_n\lambda_n\phi_{n-1}(x) = f(x)$$ This is a simple 1st order linear diff. eq solvable with integrating factors, and after solving it we repeat the process until we get to the last equation for y. Each time we replace the value of $\phi_i$ into the next equation for $\phi_{i-1}$ and solve a 1-st order eq. repetitively.

Can the "factoring" part be proven rigorously, in a similar way that it is proven for polynomials with the Fundamental Theorem of Algebra? I know I haven't given any proof for this but it seems to be true. Is this method already known? Is it any better or worse than the traditional way such equations are solved?
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 I'm sorry I forgot to add the f(x) in the beginning instead of equating with 0 (this method works for inhomogeneous equations too, otherwise it would be pointless since homogeneous eqns are easily solved anyway). I edited the original post to correct.
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Hi Boorglar! Yes, this is the characteristic equation method, see http://en.wikipedia.org/wiki/Charact...tion_(calculus)

## Solving linear differential equations by "factoring"

Hello to you!

The characteristic equation method works only when f(x) is 0 (the equation is homogeneous), but the ''factoring" method will actually work even if there is an arbitrary f(x) which makes it inhomogeneous.

Basically, I know that there is an easy method for the homogeneous case (the characteristic equation method that you showed me). But the only techniques I've heard for inhomogeneous equations involve undetermined coefficients or variation of parameters.
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor the usual method is to find a particular solution to add to the homogeneous solution wouldn't your solution involve finding n particular solutions?
 yeah it gives the particular solution together with the homogeneous one because when you integrate each 1st order eq, the constants of integration accumulate and you end up with the general homogeneous solution, plus the remaining integrals which give the particular solution. The difference is that it is more systematic than undetermined coefficients since they need some initial guess on the type of solution.

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 Quote by Boorglar … it is more systematic …
yes, but is it quicker?
 As an example this is how I would solve $y''-3y'+2y = x$ so $(D-2)(D-1)[y] = x$ $$\phi_1' - 2\phi_1 = x$$ where $\phi_1 = y'-y$ $$\phi_1(x) = e^{2x}\int{xe^{-2x}dx} + C_1e^{2x}$$ (multiply by $e^{-2x}$ (integrating factor) and write as derivative of product, etc.) $$y' - y = e^{2x}\int{xe^{-2x}dx} + C_1e^{2x}$$ So finally by muliplying by $e^{-x}$ both sides (integrating factor), etc. $$y = e^x\int e^{-x}(e^{2x}\int{xe^{-2x}dx})dx + e^x\int {C_1e^x dx} + C_2e^x$$ $$y = e^x\int (e^{x}\int{xe^{-2x}dx})dx + C_1e^{2x} + C_2e^x$$ Which is slightly ugly but gives an explicit solution for both the particular and the general solutions. This could have been done using undetermined coefficients but you'd have to guess how the function looks like before. And suppose instead of x, it was something like $y''-3y'+2y = \arctan(x^3-e^x)$. Then this would be hopeless. My method would at least be able to write it as an (albeit horribly complicated ;)) integral.

 Quote by Boorglar Hello to you! The characteristic equation method works only when f(x) is 0 (the equation is homogeneous), but the ''factoring" method will actually work even if there is an arbitrary f(x) which makes it inhomogeneous. Basically, I know that there is an easy method for the homogeneous case (the characteristic equation method that you showed me). But the only techniques I've heard for inhomogeneous equations involve undetermined coefficients or variation of parameters.
The Laplace transform technique is an established technique that is related to the characteristic equation method, is systematic, and can handle inhomogenous equations naturally without having to guess the form of the particular solution. It does not use undetermined coefficients or variation of parameters. Another related approach is the Green's function / fundamental solution, which lets you write the particular solution as a convolution integral. Both approaches would give you an integral for the particular solution that is close to the form you presented.

I think the idea you present can be made rigorous through the system of equations approach to linear constant coefficient ODEs. The system of equations you suggest can be written in the form

$\Phi' = A\Phi + b$

Where $\Phi$ is a vector of functions $\phi_n$, $A$ is a matrix of coefficients, the prime is differentiation of each component of the vector, and $b$ is a vector of all zeros, except with an $f(x)$ in one position. You cleverly chose how to define your variables so that the matrix $\Phi$ is of a nice form. However, no matter how you chose your variables, if you write the single equation as a system like above, and then bring the system to this nice form, you'll end up with your technique. Again, the characteristic equation will come up when you go to find the eigenvectors and eigenvalues of the matrix. As an example, the usual way to do this is to call $\phi_n=y^{(n)}$. Incidentally this is related to state-space methods from controls engineering too, you might want to have a look in that area for related concepts. For a start, if you google "HANDOUT ON FORMULATION OF DIFFERENTIAL EQUATIONS" and check out the thrid page of the first result, you can see why this technique works.

Also, keep in mind that linear constant coefficient equations admit complex eigenvalues, so that in your factoring, a pair of $\lambda_i$ might come in a complex conjugate pair, with corresponding sin and cosine solutions, not just real exponentials. The usual way to write these terms is to take two conjugate terms and group them into a quadratic factors, then write the rest of the monomials with real roots.
 As a side note, I believe what you propose is basically the jordan canonical form of a constant coefficient linear system.

 Tags differential, factoring, linear, operator equation