We learnd Laplace transformation, which can be used to solve diff eq.

Let $f(t)$ be a function defined for $t>0$, and we recall the following $F(p) = [LT(f)] (p) = \int_0^\infty f(t) e^{-pt} dt.$ and the inverse Laplace transformation is $f(t) = (1/2\pi i) \int_{c-i \infty}^{c+i \infty} F(p) e^{pt} dp.$ for $c \gg 1$.

For derivatives, we have $LT(Df) = p LT(f) - f(0) = p F(p) - f(0).$ and we can repeatedly use it to get $LT(DDf) = p LT(Df) - Df(0) = p (p LT(f) - f(0)) - Df(0) = p^2 F(p) - p f(0) - f'(0).$

solve equation $(D-1)(D-2) f(t) = 0$ with condition $f(0) = 0, f'(0) = 1.$

We may apply Laplace transform to the equation and get $LT [(D^2 - 3D + 2) f] = 0$ which says $p^2 F(p) - p f(0) - f'(0) - 3 [p F(p) - f(0)] + 2 F(p) = 0$ plug in the initial condition for $f(0), f'(0)$, we get $F(p) [p^2 - 3p + 2] = 1$ thus $F(p) = 1 / (p^2 - 3p + 2) = \frac{1}{(p-1)(p-2)}$

Now we can either look up the inverse Laplace transformation table, or do the inverse Laplace transformation integral, to get $f(t) = (1/2\pi i) \int_{c-i \infty}^{c+i \infty} F(p) e^{pt} dp = \sum_{p} Res_p( F(p) e^{pt})$ we have two poles,

• one is at $p=1$, with residue $e^{1 t} /(1-2) = - e^t$ and
• another at $p=2$ with residue $e^{2 t} /(2-1) = e^{2t}$

so the answer is $f(t) = e^{2t} - e^t.$ We may check that they indeed satisfies the initial condition.

If you had a equation of the form $D f(x) = 1$ then it is not a homogeneous equation: the term on the right hand side is does not contain factor $f$. What does its solution space look like? We know it is of the form $f(x) = c + x.$

Note that, the solution space is not a vector space anymore. indeed, if you have $f_1(x)$ and $f_2(x)$ both satisfies the equation, $D f_1(x) = 1$ $D f_2(x) = 1$ then add the two equation up, we see $D (f_1(x) + f_2(x)) = 2$ so $f_1(x) + f_2(x)$ is not a solution ($1 \neq 2$).

Nonetheless, the solution space is a so called 'affine space' $V$, which means there is an associated vector space $V'$, and for any two elements $v_1, v_2 \in V$, we have their difference $v_1 - v_2 \in V'$.

In our case, the associated vector space is the solution space for the homogenous equation $V' = \{f(x) \mid Df = 0 \}$

That means, if we pick a 'base point' $v_0 \in V$, and pick a basis $e_1, \cdots, e_n$ of $V'$, and then we can express any element $v \in V$ as $v = v_0 + (c_1 e_1 + \cdots + c_n e_n).$ for some coefficients $c_i$.

Back to our problem here, any solution to the inhomogenous equation can be written as a 'particular solution' (playing the role of $v_0$ above), and plus a solution to the homogenous equation (an element of $V'$).