Method of undetermined coefficients

Suppose that $f(x)$ is the solution of the homogeneous linear differential equation. Let $D = \frac{d}{dx}$. Then

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

can be expressed as

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

Now since $f(x)$ is the solution of the homogeneous linear differential equation, there exists a polynomial $H(D)$ satisfying $H(D)f(x) = 0$. Let $y_{p}$ be the particular solution of $L(D)y = f(x)$. Then

$$H(D)L(D)y_{p} = H(D)f(x) = 0$$

Thus to find $y_{p}$, all we need to find is solutions of $H(D)L(D)y = 0$ satisfying $L(D)y = f(x)$. This method is called the method of undetermind coefficient.

$$f(x) = x^{m} \Longrightarrow H(D)f(x) = D^{m+1}x^{m} = 0$$

$$f(x) = e^{ax} \Longrightarrow H(D)f(x) = (D - a)e^{ax} = 0$$

$$f(x) = e^{ax}\cos{bx} \Longrightarrow H(D)f(x) = (D^2 - 2aD + a^2 + b^2)e^{ax}\cos{bx} = 0$$

$$f(x) = e^{ax}\sin{bx} \Longrightarrow H(D)f(x) = (D^2 - 2aD + a^2 + b^2)e^{ax}\sin{bx} = 0$$

Example 2..13   Solve the differential equation $L(y) = y^{\prime\prime} - y^{\prime} - 2y = 2e^{3x}$.

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

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

Now we find the particular solution $y_{p}$ using the method of undetermined coefficients. Let $D = \frac{d}{dx}$. Then since $f(x) = 2e^{3x}$, let $H(D) = D - 3$. Then $H(D)2e^{3x} = 2(D - 3)e^{3x} = 0$. Thus $y_{p}$ is a solution of

$$(H(D)L(D))y = 2(D-3)(D^{2}-D-2)y = 0$$

The characteristic equation of this is $2(m-3)(m^{2}-m-2) = 0$. Thus the fundamental solutions are $e^{3x},e^{-x},e^{2x}$. Since $e^{-x},e^{2x}$ satisfy $L(D)y = 0$. Thus we set $y_p$ as

$$y_{p} = Ae^{3x}$$

Now put this back into $L(D)y = 2e^{3x}$. Then

$$L(D)Ae^{3x} = (Ae^{3x})^{\prime\prime} - (Ae^{3x})^{\prime} - 2(Ae^{3x}) = 4Ae^{3x} = 2e^{3x}$$

From this, we have $A = \frac{1}{2}$. 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..14   Solve the differntial equation $L(y) = y^{\prime\prime} - y^{\prime} -2y = e^{-x}$.

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

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

We next find $y_{p}$ using the method of undetermined coefficients. Since $H(D)e^{-x} = (D + 1)e^{-x} = 0$, $y_{p}$ is a solution of

$$H(D)L(D)y = (D + 1)(D^{2} - D -2)y = (D + 1)^{2}(D - 2)y = 0$$

Then the fundamental solutions are $e^{-x},xe^{-x},e^{2x}$. But $e^{-x},e^{2x}$ are solution of $L(y) = 0$. Thus we set

$$y_{p} = Axe^{-x}$$

Put this back into $L(D)y = e^{-x}$. Then

$$L(D)Axe^{-x} = -3Ae^{-x} = e^{-x}$$

Thus,

$$y_{p} = -\frac{1}{3}xe^{-x},$$

and the general solution is

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

Example 2..15   Solve the differential equation $L(y) = y^{\prime\prime} + 2y^{\prime} = e^{x} - x^{2}$.

SOLUTION The characteristic equation of $L(y) = 0$ is $m^{2} + 2m = 0$. Then we have $m = 0,-2$. Thus, the complementary solution $y_{c}$ is

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

Now to find $y_{p}$, we find the particular solution $y_{p_{1}}$ of $L(D)y = e^{x}$ and $y_{p_{2}}$ of $L(D)y = x^{2}$. Then $y_{P}$ is given by $y_{p} = y_{p_{1}} + y_{p_{2}}$.

The particular solution $y_{p_{1}}$ of $L(D)y = e^{x}$ can be found by setting $y_{p_{1}} = Ae^{x}$. Also the particular solution $y_{p_{2}}$ of $L(D)y = x^{2}$can be found by setting

$$H(D)L(D)y = (D^{3}D(D+2))y = 0$$

Then $1,x,x^{2},x^{3},e^{-2x}$ are the fundamental solutions.But $1$ and $e^{-2x}$ are complementary solutions. Thus set

$$y_{p_{2}} = Bx + Cx^{2} + Dx^{3}$$

Now let $y_{p} = y_{p_{1}} + y_{p_{2}}$ and substitute into $L(D)y = e^{x} - x^{2}$. Then $L(D)(Ae^{x} + Bx + Cx^{2} + Dx^{3}) = 3Ae^{x} + 2C + 2B + (6D + 4C)x = e^{x} - x^{2}$. Thus $A = \frac{1}{3}, B = -\frac{1}{4}, C = \frac{1}{4}, D = -\frac{1}{6}$. Therefore, the general solution is

$$y = c_{1} + c_{2}e^{-2x} + \frac{1}{3}e^{x} - \frac{1}{4}x + \frac{1}{4}x^{2} - \frac{1}{6}x^{3}\ensuremath{\ \blacksquare}$$