Differential Operator

Let $D = \frac{d}{dx}$. Then we can write

$$L(y) = a_{n}y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_{1}y^{\prime} + a_{0}y = f(x)$$

as

$$L(D)y = (a_{n}D^{n} + a_{n-1}D^{n-1} + \cdots + a_{1}D + a_{0})y = f(x)$$

Then the operator gives the function $y$ for $L(D)y = f(x)$ is called the Inverse operator and is expressed by

$$y = \frac{1}{L(D)}f(x) = (L(D))^{-1}f(x)$$

Thus $(L(D))^{-1}f(x)$ applies to the function of $f(x)$ such that

$$L(D)(L(D)^{-1}f(x)) = f(x)$$

Basic rule

Theorem 2..10 (linearlity)   Let $\alpha ,\beta$ be sonstants, $f(x),g(x)$ be functions. Then

$$L(D)(\alpha f(x) + \beta g(x)) = \alpha L(D)f(c) + \beta L(D)g(x)$$

$$L(D)^{-1}(\alpha f(x)+ \beta g(x)) = \alpha L(D)^{-1}f(x) + \beta L(D)^{-1}g(x)$$

Proof The first one comes from the definition of $L(D)$. The second one comes from the following.

$$L(D)(\alpha L(D)^{-1}f(x) + \beta L(D)^{-1}g(x)) = \alpha f(x) + \beta g(x)$$

Theorem 2..11 (Basic Formula)   Let $a$ be a constant and $m$ be an integer. Then

$$\frac{1}{D - a}f(x) = e^{ax}\int e^{-ax}f(x)\: dx$$

$$\frac{1}{(D - a)^m}f(x) = e^{ax}\iint \cdots \int e^{-ax}f(x)\: dx \cdots dx dx\ ({\rm m}\ \mbox{repeated integrals})$$

Proof Let $(D -a)^{-1}f(x) = y$. Then $(D-a)y = f(x)$. Thus

$$\frac{dy}{dx} - ay = f(x).$$

This is a linear differential equation. Thus the integrating factor $\mu = e^{\int (-a)dx}$. Multiply the $\mu$ both sides

$$d(e^{\int (-a)dx} y) = e^{\int (-a)dx} f(x).$$

Thus

$$y = e^{-\int(-a)dx}\int e^{\int (-a)dx} f(x)\:dx = e^{ax}\int e^{-ax}f(x)\:dx$$

Repeating this

$$\frac{1}{(D-a)^2} = \frac{1}{D-a}\big(\frac{1}{D-a} f(x)\big) = e^{ax}\int e^{-ax}(\frac{1}{D-a}f(x))\:dx$$

$$= e^{ax}\int e^{-ax}(e^{ax}\int e^{-ax}f(x))\:dx$$

$$= e^{ax}\iint e^{-ax}f(x)\:dx$$

Theorem 2..12 (Differential Operator Properties)  

$$(1)\ f(x) = e^{ax} \Longrightarrow \frac{1}{L(D)}e^{ax} = \frac{1}{L(a)}e^{ax}$$

$$(2)\ f(x) = x^{m} \Longrightarrow \frac{1}{1-aD}x^{m} = \big(1 + (aD) + (aD)^2 + \cdots (aD)^m\big) x^m$$

$$(3)\ f(x) = \cos{ax} \Longrightarrow \frac{1}{L(D^2)}\cos{ax} = \frac{1}{L(-a^2)}\cos{ax}$$

$$(4)\ f(x) = \sin{ax} \Longrightarrow \frac{1}{L(D^2)}\sin{ax} = \frac{1}{L(-a^2)}\sin{ax}$$

$$(5)\ f(x) = e^{ax}u(x) \Longrightarrow \frac{1}{L(D)}e^{ax}u(x) = e^{ax}\frac{1}{L(D + a)}u(x)$$

$$(6)\ f(x) = x^m u(x) \Longrightarrow \frac{1}{L(D)}x^m u(x) = \sum_{k=0}^{m}\binom{m}{k}x^{m-k}\big(\frac{1}{L(D)}\big)^{(k)}u(x)$$

Example 2..20   Solve the differential equation $L(y) = y^{\prime\prime} - y^{\prime} - 2y = 2e^{3x}$.

SOLUTION The characteristic equation of $L(y) = 0$ is $m^{2} - m -2 = 0$. Thus roots are $m = -1,2$. Then the complementary solution $y_{c}$ is given by

$$y_{c} = c_{1}e^{-x} + c_{2}e^{2x}$$

Now we find the particular solution. Since $(D^2 - D - 2) y = 2e^{3x}$, by (1) above

$$y = \frac{2e^{3x}}{D^2 - D -1} = \frac{2 e^{3x}}{3^2 - 3 -1} = \frac{2 e^{3x}}{4} = \frac{e^{3x}}{2}.$$

Thus $y_{p} = \frac{1}{2}e^{3x}$ and the general solution is

$$y = y_{c} + y_{p} = c_{1}e^{-x} + c_{2}e^{2x} + \frac{1}{2}e^{3x}\ensuremath{\ \blacksquare}$$

Example 2..21   Solve the differential equation $L(y) = y^{\prime\prime} + 4y = \sin{x}$.

SOLUTION The characteristic equation of $L(y) = 0$ is $m^2 + 4 = 0$. Thus $m = \pm 2i$. Then the complementary solution is

$$y_{c} = c_{1}\cos{2x} + c_{2}\sin{2x}$$

Now we find the particular solution. Since $f(x) = \sin{x}$, we have $(D^2 + 4) y = \sin{x}$ and

$$y = \frac{\sin{x}}{D^2 + 4} = \frac{\sin{x}}{-1^2 + 4} = \frac{1}{3}\sin{x}.$$

Example 2..22   Find the particular solution of $L(y) = y^{\prime\prime} -3y' + 2y = xe^{x}$

SOLUTION Let $y_p$ be the particular solution. Then since $f(x) = xe^x$, we have

$$y_p = \frac{1}{D^2 - 3D +2}(xe^{x}) = e^{x}\frac{1}{(D+1)^2 -3(D+1)+2}x = e^{x}\frac{1}{D^2 - D}x$$

$$= e^{x} \frac{1}{D} \cdot \frac{1}{D-1}x = -e^{x}\frac{1}{D} \frac{1}{1-D}x$$

$$= -e^{x}\frac{1}{D} (1+D+D^2 + \cdots)x = -e^{x}\frac{1}{D}(x+1) = -e^{x}\big(\frac{1}{2}x^2 + x\big)$$