6.2 Linear Equations of the Second Order
An equation of the form \[a_0(x)\frac {d^2y}{dx^2} + a_1(x)\frac {dy}{dx} + a_2(x)y = F(x) \hspace {0.3cm}\cdots \cdots \cdots \hspace {0.4cm} (12)\] is a linear differential equation of the second order. It is called homogeneous if \(F(x) = 0\) and non-homogeneous otherwise. The word homogeneous is being used here in a completely different sense from the first-order equations above, where it referred to the degree of \(M\) and \(N\); this is unfortunate but standard, and the meaning is always clear from the order of the equation.
Superposition of solutions
If \(y_1\) and \(y_2\) are solutions of the homogeneous equation, so is \(Ay_1 + By_2\) for any constants \(A\) and \(B\). This follows immediately from linearity: substituting \(Ay_1 + By_2\) into the left-hand side of \((12)\) and collecting terms gives \(A\) times the result for \(y_1\) plus \(B\) times the result for \(y_2\), and both are zero.
If in addition \(y_1\) and \(y_2\) are linearly independent — neither is a constant multiple of the other — then \[y_c = Ay_1 + By_2\] contains every solution of the homogeneous equation, and is called the complementary function.
The general solution of the non-homogeneous equation
Suppose \(y_p\) is any single solution of \((12)\), called a particular integral. If \(y\) is any other solution then \(y - y_p\) satisfies the homogeneous equation, because the two \(F(x)\) terms cancel; so \(y - y_p = y_c\) for some choice of constants. Therefore every solution has the form \[\boxed {y = y_c + y_p}\qquad \text {that is}\qquad GS = CF + PI .\] This is the organising fact for everything that follows: the work splits into finding the complementary function, which needs only the left-hand side, and finding one particular integral, which is where \(F(x)\) enters.
6.2.1 Homogeneous Equations with Constant Coefficients
Take \[a\frac {d^2y}{dx^2} + b\frac {dy}{dx} + cy = 0 ,\] with \(a, b, c\) constants and \(a\neq 0\). Trying \(y = e^{mx}\), which is plausible because the exponential reproduces itself on differentiation, gives \[\big (am^2 + bm + c\big )e^{mx} = 0 .\] Since \(e^{mx}\) is never zero, \(y = e^{mx}\) is a solution precisely when \[\boxed {am^2 + bm + c = 0}\] which is called the auxiliary equation. Its roots determine everything, and there are three cases.
Case 1: real distinct roots \(m_1 \neq m_2\)
\[y = Ae^{m_1x} + Be^{m_2x}\]
Case 2: a repeated real root \(m\)
Here \(e^{mx}\) supplies only one solution. The second is \(xe^{mx}\), and \[y = \big (A + Bx\big )e^{mx}\]
Case 3: complex roots \(m = \alpha \pm i\beta \)
The solution \(Ae^{(\alpha +i\beta )x} + Be^{(\alpha -i\beta )x}\) is rewritten in real form using Euler’s relation \(e^{i\theta } = \cos \theta + i\sin \theta \), giving \[y = e^{\alpha x}\big (A\cos \beta x + B\sin \beta x\big )\]
Example 6.2.1. Solve the following.
- 1.
- \(\dfrac {d^2y}{dx^2} - 5\dfrac {dy}{dx} + 6y = 0\)
- 2.
- \(\dfrac {d^2y}{dx^2} - 6\dfrac {dy}{dx} + 9y = 0\)
- 3.
- \(\dfrac {d^2y}{dx^2} + 4\dfrac {dy}{dx} + 13y = 0\)
Solution.
Part 1
The auxiliary equation is \[m^2 - 5m + 6 = 0 \implies (m-2)(m-3) = 0 \implies m = 2, 3 ,\] real and distinct, so \[y = Ae^{2x} + Be^{3x}.\]
Part 2
\[m^2 - 6m + 9 = 0 \implies (m-3)^2 = 0 \implies m = 3 \text { twice},\] a repeated root, so \[y = \big (A + Bx\big )e^{3x}.\]
Part 3
\[m^2 + 4m + 13 = 0 \implies m = \frac {-4 \pm \sqrt {16 - 52}}{2} = \frac {-4 \pm 6i}{2} = -2 \pm 3i ,\] so \(\alpha = -2\) and \(\beta = 3\), giving \[y = e^{-2x}\big (A\cos 3x + B\sin 3x\big ).\]
6.2.2 The Method of Undetermined Coefficients
To find a particular integral when \(F(x)\) is a polynomial, an exponential, a sine or cosine, or a sum or product of these, assume \(y_p\) has the same form with unknown coefficients, substitute, and match.
| \(F(x)\) | try \(y_p\) |
| polynomial of degree \(n\) | polynomial of degree \(n\) |
| \(ke^{rx}\) | \(Ce^{rx}\) |
| \(k\cos \omega x\) or \(k\sin \omega x\) | \(C\cos \omega x + D\sin \omega x\) |
Note. If the trial form already appears in the complementary function it cannot work, since it makes the left-hand side zero. Multiply the trial solution by \(x\) — and again by \(x\) if necessary — until it no longer duplicates any term of the complementary function. The second example below is exactly this case.
Solution. The complementary function was found above: \(y_c = Ae^{2x} + Be^{3x}\).
Since \(e^{x}\) does not appear in \(y_c\), try \(y_p = Ce^{x}\). Then \(y_p' = Ce^{x}\) and \(y_p'' = Ce^{x}\), so substituting, \[Ce^{x} - 5Ce^{x} + 6Ce^{x} = 2e^{x} \implies 2Ce^{x} = 2e^{x} \implies C = 1 .\] Hence \(y_p = e^{x}\) and \[\boxed {y = Ae^{2x} + Be^{3x} + e^{x}}\]
Solution. The complementary function is again \(y_c = Ae^{2x} + Be^{3x}\), and this time \(e^{2x}\) is one of its terms. Trying \(y_p = Ce^{2x}\) would give zero on the left, so multiply by \(x\) and try \[y_p = Cxe^{2x}.\] Then \[y_p' = C\big (1 + 2x\big )e^{2x},\qquad y_p'' = C\big (4 + 4x\big )e^{2x},\] and substituting, \[C\Big [\big (4 + 4x\big ) - 5\big (1 + 2x\big ) + 6x\Big ]e^{2x} = e^{2x}.\] The bracket simplifies to \(4 + 4x - 5 - 10x + 6x = -1\), so \(-C = 1\) and \(C = -1\). Hence \[\boxed {y = Ae^{2x} + Be^{3x} - xe^{2x}}\] Note that the \(x\) terms cancelled identically. That is not luck: it happens precisely because \(e^{2x}\) solves the homogeneous equation, and it is the check that the trial form was the right one.
6.2.3 The Method of Variation of Parameters
Undetermined coefficients requires \(F(x)\) to be of a special form. Variation of parameters has no such restriction, at the cost of two integrals.
Given the complementary function \(y_c = Ay_1 + By_2\), look for a particular integral of the same shape but with the constants replaced by functions: \[y_p = u_1(x)y_1 + u_2(x)y_2 .\] Imposing the convenient extra condition \(u_1'y_1 + u_2'y_2 = 0\) and substituting into the equation \(y'' + py' + qy = F\) leads to the pair \[u_1'y_1 + u_2'y_2 = 0,\qquad u_1'y_1' + u_2'y_2' = F ,\] whose solution is \[\boxed {u_1' = \frac {-y_2F}{W},\qquad u_2' = \frac {y_1F}{W}}, \qquad \text {where}\qquad W = \begin {vmatrix} y_1 & y_2\\ y_1' & y_2'\end {vmatrix}\] is the Wronskian of \(y_1\) and \(y_2\). It is non-zero precisely because \(y_1\) and \(y_2\) are linearly independent.
Solution. The auxiliary equation \(m^2 + 1 = 0\) has roots \(m = \pm i\), so \[y_c = A\cos x + B\sin x ,\qquad y_1 = \cos x,\quad y_2 = \sin x .\] The Wronskian is \[W = \begin {vmatrix} \cos x & \sin x\\ -\sin x & \cos x\end {vmatrix} = \cos ^2x + \sin ^2x = 1 .\] Note that undetermined coefficients is unavailable here: \(\sec x\) is not a polynomial, exponential or sinusoid. Using the formulae, \[u_1' = \frac {-y_2F}{W} = -\sin x\sec x = -\tan x \implies u_1 = \ln \left |\cos x\right | ,\] \[u_2' = \frac {y_1F}{W} = \cos x\sec x = 1 \implies u_2 = x .\] Therefore \[y_p = \cos x\ln \left |\cos x\right | + x\sin x\] and \[\boxed {y = A\cos x + B\sin x + \cos x\ln \left |\cos x\right | + x\sin x}\]
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