I2, I4, I6 - Auxiliary equations with real roots | AQA A-Level Further Mathematics

A second-order differential equation links a function to both its slope and the rate at which its slope changes. I4 uses an auxiliary quadratic to find exponential solutions; I6 explains why distinct and repeated roots give different forms. I2 then uses two initial conditions to select a particular solution.

Distinct and repeated real roots

Can a curve start at $y=1$ with horizontal tangent and still tend to zero while satisfying $y''+4y'+4y=0$ ? A single decaying exponential has a nonzero slope when its value is one, so we will need more than one adjustable constant.

Here $y'=\mathrm{d}y/\mathrm{d}x$ and $y''=\mathrm{d}^2y/\mathrm{d}x^2$ . “Second order” refers to the highest derivative, not a square of the first derivative. The equation is homogeneous because its right side is zero.

Try $y=e^{mx}$ in $y''+ay'+by=0$ , where $a,b$ are real constants. Exponentials are useful because differentiation preserves their shape:

\begin{aligned} y'&=me^{mx},\qquad y''=m^2e^{mx}\\ y''+ay'+by&=(m^2+am+b)e^{mx} \end{aligned}

Since $e^{mx}\ne0$ , this is zero exactly when $m^2+am+b=0$ .

Auxiliary equation

The auxiliary equation of $y''+ay'+by=0$ , with constant coefficients, is

m^2+am+b=0

Its roots identify exponential solution modes. If the original coefficient of $y''$ is a nonzero constant other than one, divide through first, or use the corresponding unnormalised quadratic.

Worked example 1: two distinct roots

Question. Find the auxiliary roots for $y''+4y'+3y=0$ , and identify the corresponding exponential solutions.

Plan. Form the quadratic with the same coefficients and factorise it.

\begin{aligned} m^2+4m+3&=0\\ (m+1)(m+3)&=0\\ m&=-1\quad\text{or}\quad m=-3 \end{aligned}

Answer. The roots are distinct: $-1,-3$ . Corresponding solutions are $e^{-x}$ and $e^{-3x}$ .

Check. Substitution of $e^{-x}$ gives the coefficient $1-4+3=0$ ; substitution of $e^{-3x}$ gives $9-12+3=0$ .

Now change only the coefficient of $y$ from $3$ to $4$ . The two roots merge.

Worked example 2: a repeated root

Question. Find the auxiliary roots for $y''+4y'+4y=0$ , and identify its exponential solution.

Plan. Factorise the new quadratic and retain the root's multiplicity.

\begin{aligned} m^2+4m+4&=0\\ (m+2)^2&=0\\ m&=-2\quad\text{repeated} \end{aligned}

Answer. The root is $-2$ , repeated twice; $e^{-2x}$ is a solution. Finding the root twice does not create two independent exponentials.

Check. For $y=e^{-2x}$ , $y''+4y'+4y=(4-8+4)e^{-2x}=0$ .

The same exponential written twice only combines into one coefficient. The next part shows how to obtain the missing independent solution.

Watch out

Wrong: replacing $y''$ by $2m$ . Differentiating $e^{mx}$ twice produces $m^2e^{mx}$ , so the auxiliary equation is quadratic in $m$ .

Identify a fresh repeated root before choosing a general solution.

General solutions for real roots

Because the equation is linear and homogeneous, multiplying a solution by a constant or adding two solutions gives another solution. Two independent solution modes provide two genuinely adjustable constants.

Real-root solution forms

For $y''+ay'+by=0$ , with real constant coefficients:

Distinct real roots $r_1,r_2$ give $y=Ae^{r_1x}+Be^{r_2x}$ .
A repeated real root $r$ gives $y=(A+Bx)e^{rx}$ .

In each case $A,B$ are arbitrary real constants. These are general solutions, including the zero solution.

Not in the formula booklet: learn it.

The repeated-root factor $x$ is essential, and comes from integration rather than a memorised repair. A repeated root means $a=-2r$ , $b=r^2$ . Put $y=v(x)e^{rx}$ . Product differentiation gives

y''-2ry'+r^2y=e^{rx}v''

Thus $v''=0$ , so $v=A+Bx$ . This proves the repeated-root form and supplies two independent solutions $e^{rx}$ , $xe^{rx}$ .

For distinct roots, completeness can also be traced to first-order equations. Set $z=y'-r_2y$ . Since $r_1+r_2=-a$ and $r_1r_2=b$ , the original equation becomes $z'-r_1z=0$ . Hence $z=Ke^{r_1x}$ . Solving $y'-r_2y=Ke^{r_1x}$ with integrating factor $e^{-r_2x}$ gives the two distinct exponentials. This connects the new method directly to integrating factors.

Worked example 3: combining distinct modes

Question. Find the general solution of $y''+4y'+3y=0$ .

Plan. Use the two distinct auxiliary roots and give each independent mode its own constant.

\begin{aligned} m^2+4m+3&=(m+1)(m+3)=0\\ y&=Ae^{-x}+Be^{-3x} \end{aligned}

Answer. $y=Ae^{-x}+Be^{-3x}$ , for real $x$ .

Check. The residual is $A(1-4+3)e^{-x}+B(9-12+3)e^{-3x}=0$ for all $A,B$ . Also the two exponentials cannot be constant multiples of one another on an interval, since their ratio is $e^{2x}$ .

Again change only the coefficient of $y$ from $3$ to $4$ . Repeating the root changes the structure of the general solution, not the number of constants needed.

Worked example 4: supplying the missing independent solution

Question. Find the general solution of $y''+4y'+4y=0$ .

Plan. Recognise the repeated root and use a linear factor multiplying its exponential.

\begin{aligned} (m+2)^2&=0\\ y&=(A+Bx)e^{-2x}\\ y'&=(B-2A-2Bx)e^{-2x}\\ y''&=(4A-4B+4Bx)e^{-2x} \end{aligned}

Answer. $y=(A+Bx)e^{-2x}$ , for real $x$ .

Check. In $y''+4y'+4y$ , the coefficients inside the common exponential sum to

(4A-4B+4Bx)+(4B-8A-8Bx)+(4A+4Bx)=0

The ratio $(xe^{-2x})/(e^{-2x})=x$ is not constant, so the two modes are independent.

If the repeated root is zero, the same rule gives $y=A+Bx$ , exactly the result of integrating $y''=0$ twice. A zero root is a valid root, not a reason to discard a mode.

Watch out

Wrong: $y=Ae^{-2x}+Be^{-2x}$ as the general solution for a repeated root. It reduces to $(A+B)e^{-2x}$ , with only one effective constant. Use $(A+Bx)e^{-2x}$ .

Apply the repeated-root rule with a new root.

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I2, I4, I6 - Auxiliary equations with real roots

Distinct and repeated real roots

General solutions for real roots

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