Constructing the Particular Solution from Complementary Solution
Another more general method to derive the particular solution for a non-homogeneous second-order ODE is the Variation of Parameters. Assume that the complementary solution takes the form of \( y_c = c_1 y_1 + c_2 y_2 \) where \( y_1(x), y_2(x) \) are two linearly independent bases. The essence of this method is to “vary” the constants \( c_1, c_2 \) so now they are some parameters \( p_1(x), p_2(x) \) that are functions of \(x\), and the particular solution will look like \( y_p = p_1(x) y_1 + p_2(x) y_2 \). The corresponding first derivative is
\begin{align}
y_p’ &= p_1^\prime y_1 + p_1 y_1^\prime + p_2^\prime y_2 + p_2 y_2^\prime \\
&= (p_1^\prime y_1 + p_2^\prime y_2) + p_1 y_1^\prime + p_2 y_2^\prime \\
&= p_1 y_1^\prime + p_2 y_2^\prime \tag{1}
\end{align}
where we have made a very important assumption that
\begin{equation}
p_1^\prime y_1 + p_2^\prime y_2 = 0 \tag{2}
\end{equation}
This will be of use later. Subsequently, the second derivative will be
\begin{align}
y_p^{\prime\prime} = p_1^\prime y_1^\prime + p_1 y_1^{\prime\prime} + p_2^\prime y_2^\prime + p_2 y_2^{\prime\prime} \tag{3}
\end{align}
For the original second-order non-homogeneous ODE:
\begin{equation}
\frac{d^2y}{dx^2} + \alpha\frac{dy}{dx} + \beta y = G(x) \tag{4}
\end{equation}
where \( G(x) \) is the source term, substituting the particular solution with (1) and (3) leads to
\begin{align}
y_p^{\prime\prime} + \alpha y_p’ + \beta y_p &= G(x) \\
\begin{aligned}
&p_1^\prime y_1^\prime + p_1 y_1^{\prime\prime} + p_2^\prime y_2^\prime + p_2 y_2^{\prime\prime} \\
& + \alpha (p_1 y_1^\prime + p_2 y_2^\prime) + \beta (p_1 y_1 + p_2 y_2)
\end{aligned}
&= G(x) \\
\begin{aligned}
&p_1 y_1^{\prime\prime} + \alpha p_1 y_1^\prime + \beta p_1 y_1 + p_2 y_2^{\prime\prime} + \alpha p_2 y_2^\prime + \beta p_2 y_2 \\
& + p_1^\prime y_1^\prime + p_2^\prime y_2^\prime
\end{aligned}
&= G(x) \\
\begin{aligned}
&p_1 (y_1^{\prime\prime} + \alpha y_1^\prime + \beta y_1) + p_2 (y_2^{\prime\prime} + \alpha y_2^\prime + \beta y_2) \\
& + p_1^\prime y_1^\prime + p_2^\prime y_2^\prime
\end{aligned}
&= G(x) \\
p_1^\prime y_1^\prime + p_2^\prime y_2^\prime &= G(x) \tag{5}
\end{align}
The bracketed terms vanish because \( y_1, y_2 \) are exactly the complementary solutions that satisfy the original ODE: \( y_1^{\prime\prime} + \alpha y_1^\prime + \beta y_1 = 0\) and \( y_2^{\prime\prime} + \alpha y_2^\prime + \beta y_2 = 0 \). Equations (2) and (5) then form a linear system
\begin{align}
\left\{\begin{aligned}
p_1^\prime y_1 + p_2^\prime y_2 &= 0 \\
p_1^\prime y_1^\prime + p_2^\prime y_2^\prime &= G(x)
\end{aligned}\right. \tag{6}
\end{align}
which can be solved for \( p_1^\prime, p_2^\prime \) (\(y_1, y_2\) are known). Integrating them respectively then yields \( p_1, p_2 \) and hence we shall retrieve the particular solution.
Example
Find the particular solution for the non-homogeneous ODE:
\begin{equation}
\frac{d^2y}{dx^2} + 4y = \tan x \tag{7}
\end{equation}
The two complementary solutions are easily seen to be \( y_1 = \cos(2x) \) and \( y_2 = \sin(2x) \). Now we let \( y_p = v_1(x)y_1 + v_2(x)y_2 \) and need to solve
\begin{align}
\left\{\begin{aligned}
v_1’y_1 + v_2’y_2 &= 0 \\
v_1’y_1^\prime + v_2’y_2^\prime &= \tan x
\end{aligned}\right.
\Rightarrow
\left\{\begin{aligned}
v_1’\cos(2x) + v_2’\sin(2x) &= 0 \\
v_1^\prime(-2\sin(2x)) + v_2^\prime(2\cos(2x)) &= \tan x
\end{aligned}\right. \tag{8}
\end{align}
according to (6) with \( p \) replaced by \( v \). By Cramer’s Rule in Linear Algebra, we get
\begin{align}
v_1^\prime &= \frac{
\begin{vmatrix}
0 & \sin(2x) \\
\tan x & 2\cos(2x)
\end{vmatrix}
}{W} \\
&= \frac{1}{2}(-\sin(2x)\tan(x)) \\
&= -\frac{1}{2}(2\sin x\cos x)\frac{\sin x}{\cos x} = -\sin^2 x \tag{9}
\end{align}
where the Wronskian is
\begin{align}
W[y_1, y_2] &=
\begin{vmatrix}
\cos(2x) & \sin(2x) \\
-2\sin(2x) & 2\cos(2x)
\end{vmatrix} \\
&= 2\cos^2(2x) + 2\sin^2(2x) = 2 \tag{10}
\end{align}
Similarly,
\begin{align}
v_2^\prime &= \frac{1}{2}\cos(2x)\tan x \\
&= \frac{1}{2}(2\cos^2 x -1)\tan x \\
&= \sin x \cos x -\frac{1}{2} \tan x = \frac{1}{2} \sin 2x -\frac{1}{2} \tan x \tag{11}
\end{align}
Now what remains is to integrate them to obtain
\begin{align}
v_1 &= \int -\sin^2 x dx \\
&= \int \frac{\cos (2x) -1}{2} dx = \frac{1}{4} \sin (2x) -\frac{1}{2}x \tag{12} \\
v_2 &= \int (\frac{1}{2} \sin 2x -\frac{1}{2} \tan x) dx \\
&= -\frac{1}{4} \cos (2x) + \frac{1}{2} \ln (\cos x) \tag{13}
\end{align}
(No need to include the integration constants, why?) and thus
\begin{align}
y_p &= v_1y_1 + v_2y_2 \\
&= [\frac{1}{4} \sin (2x) -\frac{1}{2}x](\cos(2x)) \\
&\quad+ [-\frac{1}{4} \cos (2x) + \frac{1}{2} \ln (\cos x)](\sin(2x)) \\
&= -\frac{1}{2}x\cos(2x) + \frac{1}{2}\sin(2x)\ln (\cos x) \tag{14}
\end{align}
where we have absorbed the terms same as the complementary solutions.
Exercise
Derive the full solution of
\begin{equation}
2\frac{d^2y}{dx^2} -3\frac{dy}{dx} + y = xe^x \tag{15}
\end{equation}
where \( y(0) = 1, y'(0) = 0\), by Variation of Parameters.
Answer
The auxiliary equation is seen to be \( 2r^2 -3r + 1 = 0 \) which has the root of \( r = 1/2, 1 \). Therefore, the complementary solutions are \( y_1 = e^{x/2} \) and \( y_2 = e^x \). We then seek to solve (6); however, note that the leading coefficient of (15) is \( 2 \), so we need to normalize it by multiplying \( 1/2 \), and \(G(x)\) will be \( xe^x/2 \). The linear system is then
\begin{align}
\left\{\begin{aligned}
p_1’y_1 + p_2’y_2 &= 0 \\
p_1’y_1^\prime + p_2’y_2^\prime &= xe^x/2
\end{aligned}\right.
\Rightarrow
\left\{\begin{aligned}
p_1’e^{x/2} + p_2’e^{x} &= 0 \\
p_1^\prime((1/2)e^{x/2}) + p_2^\prime e^{x} &= xe^x/2
\end{aligned}\right.
\end{align}
This is not very hard to solve: \( p_1^\prime = -xe^{x/2} \) and \( p_2^\prime = x \), and
\begin{align}
p_1 &= -\int xe^{x/2} dx \\
&= -2\int x d(e^{x/2}) \\
&= -2[xe^{x/2}] + 2\int e^{x/2}dx \\
&= -2xe^{x/2} + 4e^{x/2}
\end{align}
also \( p_2 = x^2/2\) easily. The particular solution is hence
\begin{align}
y_p &= p_1y_1 + p_2y_2 \\
&= (-2xe^{x/2} + 4e^{x/2})e^{x/2} + (x^2/2)e^x \\
&= 4e^x -2xe^x + x^2e^x/2 = -2xe^x + x^2e^x/2
\end{align}
where the \(4e^x\) term can be absorbed into \( y_2 \). The general solution will be
\begin{align}
y = c_1e^{x/2} + c_2e^x -2xe^x + x^2e^x/2
\end{align}
and \( y’ = (c_1/2)e^{x/2} + (c_2-2)e^x -xe^x + x^2e^x/2 \). Matching the I.C. reveals that \( c_1 = -2, c_2 = 3 \), and the full solution is therefore
\begin{equation}
y = -2e^{x/2} + (x^2/2 -2x +3)e^x
\end{equation}
This question can also be solved by the previous method of Undetermined Coefficients.
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