Today we considered solving homogenous constant coefficient differential equation.

In general, the equation you meet looks like $$ (d/dx)^n f(x) + c_{n-1} (d/dx)^{n-1} f(x) + \cdots + c_1 (d/dx) f(x) + c_0 f(x) = 0. $$ If we call $D = d/dx$ and factor out $f(x)$ on the right, we can write the above equation as $$ (D^n + c_{n-1} D^{n-1} + \cdots + c_0 ) f(x) = 0. $$ We sometimes use $P(D) = D^n + c_{n-1} D^{n-1} + \cdots + c_0$, $P(D)$ is a degree $n$ polynomial in $D$.

Let $V$ denote the set of solutions for the equation. Because the equation is homogeneous in $f$ (meaning, each term in the equation has one and only one factor of $f$), the solution space is a vector space, meaning you can add two solutions together and still get a solution.

If $P(D)$ is a degree $n$ operator, then the solution space is $n$ dimensional.

Here we try to find general solution for the equation $P(D) f(x) = 0$.

As long as you can find $n$ linearly independent solutions, call them $f_1(x), \cdots, f_n(x)$, then you win. Since they will form a basis of the solution space.

We factorize $P(D)$ into linear factors $$ P(D) = (D - \lambda_1) \cdots (D - \lambda_n) $$ we can always do this by the 'fundamental theorem of algebra', which says any polynomials admits such factorization.

Suppose all $\lambda_i$ are distinct, then the following is a list of $n$ linearly independent solutions $$ e^{\lambda_1 x}, \cdots, e^{\lambda_n x} $$

It is important to note that, $D$ does not 'commute' with $x$ $$ D (x f(x)) \neq x D f(x). $$ but $D$ commute with itself, $$ (D-a) (D-b) f(x) = (D-b) (D-a) f(x). $$

What if we had repeated roots?

See Monday's example.

If you face an equation like this $$ D^k f(x) = 0. $$ then you know you can have a basis of solution like $1, x, \cdots, x^{k-1}$.

What about $$ (D-\lambda)^k f(x) = 0 ?$$

We introduced a trick $$ D (e^{ax} f(x)) = (a e^{ax} f(x) + e^{ax} D f(x)) = e^{ax} (D+a) f(x). $$ hence $$ (D-a) (e^{ax} f(x)) = e^{ax} D f(x). $$ or a variant $$ e^{-ax} D (e^{ax} f(x)) = (D+a) f(x). $$

We can write $f(x) = e^{\lambda x} g(x)$, then we have $$ (D-\lambda)^k f(x) = (D-\lambda)^k [e^{\lambda x} g(x)] = (D-\lambda)^{k-1} [e^{\lambda x} D g(x)] = (D-\lambda)^{k-2} [e^{\lambda x} D^2 g(x)] = e^{\lambda x} D^k g(x) $$ now, we know the equation for $g(x)$ is $D^k g(x) = 0$, and we know the general solution for $g(x)$ is $$ g(x) = c_0 + c_1 x + \cdots + c_{k-1} x^{k-1}. $$ hence general solution for $f(x)$ is $$ f(x) = e^{\lambda x} (c_0 + c_1 x + \cdots + c_{k-1} x^{k-1}) $$

In general, we can write $$ P(D) = (D - \lambda_1)^{m_1} \cdots (D - \lambda_r)^{m_r} $$ where $\lambda_i$ are distinct, and the multiplicity $m_1, \cdots, m_r$ add up to $n$. Then we have the following general solutions $$ f(x) = \sum_{i=1}^r \sum_{j=0}^{m_i-1} c_{i,j} x^j e^{\lambda_i x}. $$