2.6 - Complementary functions | Cambridge A-Level Further Mathematics

A linear differential equation separates into a response to its right-hand side and a family of solutions to the corresponding zero-right-hand-side equation. You will combine these parts and find the complementary function when the auxiliary roots are distinct, repeated or complex.

Homogeneous and particular parts

Why can two different functions both satisfy $y'+2y=6$ ? The constant function $y=3$ works, but adding $e^{-2x}$ still works because its derivative cancels twice the added function.

Complementary function and particular integral

For a linear equation $L[y]=f(x)$ , the complementary function $y_c$ is the general solution of the associated homogeneous equation $L[y]=0$ . A particular integral $y_p$ is any one solution of $L[y]=f(x)$ .

The general solution is $y=y_c+y_p$ . A particular integral contains no arbitrary constants that need to be added to the complementary function.

Here $L$ means the given linear combination of $y$ and its derivatives. Linearity gives $L[y_c+y_p]=L[y_c]+L[y_p]=0+f$ . Conversely, subtracting any particular integral from any solution gives a homogeneous solution. This proves that the sum describes the whole family, rather than just some solutions.

A particular integral need not be unique: adding a fixed homogeneous solution produces another one. The resulting general family is unchanged. A particular solution usually means the member selected by specified initial conditions; it is not the same term as particular integral.

Worked example 1: combining a first-order family

Question. For $y'+2y=6$ , verify the particular integral $y_p=3$ and use the given complementary function $y_c=Ce^{-2x}$ to state the general solution.

Plan. Substitute the particular integral, then add the homogeneous family.

For $y_p=3$ , its derivative is zero, so $y_p'+2y_p=0+6=6$ . The complementary term contributes zero to the left-hand side.

Answer. $y=3+Ce^{-2x}$ .

Check. Solving instead by separation gives $\mathrm{d}(y-3)/\mathrm{d}x=-2(y-3)$ , hence $y-3=Ce^{-2x}$ , including the equilibrium solution when $C=0$ .

Now change the order of the equation. A second-order equation needs a two-constant complementary function.

Worked example 2: combining a second-order family

Question. For $y''+2y'=6$ , verify $y_p=3x$ . The complementary function is $y_c=A+Be^{-2x}$ . State the general solution.

Plan. Check the supplied particular integral and retain both arbitrary constants.

Since $y_p'=3$ and $y_p''=0$ , the particular integral gives $0+2(3)=6$ .

Answer. $y=A+Be^{-2x}+3x$ .

Check. Put $v=y'$ in the original equation. Then $v'+2v=6$ , so the preceding first-order result gives $v=3+Ke^{-2x}$ . Integration gives $y=3x-(K/2)e^{-2x}+A$ , the same family after renaming the constant.

The second example's constant $A$ cannot be dropped: differentiation removes it, so the differential equation cannot determine it.

Watch out

A particular integral alone is one solution. The general solution must include the complementary function and its full set of independent arbitrary constants.

Combine the two supplied pieces in another equation.

Distinct and repeated auxiliary roots

An exponential is a useful trial for a constant-coefficient homogeneous equation because differentiation only multiplies it by a constant.

For example, substituting $y=e^{mx}$ in $y'+3y=0$ gives $(m+3)e^{mx}=0$ . Since the exponential is nonzero, $m=-3$ , giving the first-order complementary function $Ce^{-3x}$ .

For $ay''+by'+cy=0$ , where $a\ne0$ , the same substitution gives

(am^2+bm+c)e^{mx}=0.

The auxiliary equation is therefore $am^2+bm+c=0$ .

Real auxiliary roots

For a second-order homogeneous linear equation with constant coefficients:

\begin{array}{ll} m_1\ne m_2\text{ real}: & y_c=Ae^{m_1x}+Be^{m_2x} \\ m=r\text{ repeated}: & y_c=(A+Bx)e^{rx}. \end{array}

For a first-order equation $ay'+by=0$ , where $a\ne0$ , the complementary function is $Ce^{-bx/a}$ . These results are not in the formula booklet: learn them.

Why does a repeated root require $x$ ? Writing $Ae^{rx}+Be^{rx}$ produces only one effective constant. For the repeated-root equation $y''-2ry'+r^2y=0$ , substitution of $y=e^{rx}v$ leaves $e^{rx}v''=0$ . Thus $v=A+Bx$ , supplying the missing independent solution.

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2.6 - Complementary functions

Homogeneous and particular parts

Distinct and repeated auxiliary roots

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