4.10d - Second order homogeneous equations | OCR A-Level Further Mathematics A

A quadratic equation can determine an entire family of functions. You will use the auxiliary equation to solve second order homogeneous differential equations, choose the right form for real, repeated or complex roots, and determine the constants from conditions.

The auxiliary equation and distinct roots

Why can solving a quadratic tell you the shape of a curve? Exponentials keep their form when differentiated: if $y=e^{mx}$ , then $y'=me^{mx}$ and $y''=m^2e^{mx}$ . This turns a differential equation into algebra.

You already know

A second order general solution has two independent arbitrary constants. A quadratic has distinct real roots, a repeated real root or conjugate complex roots according as its discriminant is positive, zero or negative.

A second order linear equation with constant coefficients can be written $y''+ay'+by=0$ , where $a$ and $b$ are real constants. Homogeneous means that its right-hand side is zero; it does not mean that every coefficient is zero. If the coefficient of $y''$ is a non-zero constant, divide by it first.

Substitute the trial function $e^{mx}$ :

\begin{aligned} m^2e^{mx}+ame^{mx}+be^{mx}&=0 &&\text{substitute the derivatives}\\ e^{mx}(m^2+am+b)&=0 &&\text{factor out }e^{mx}\\ m^2+am+b&=0 && e^{mx}\ne0 \end{aligned}

Auxiliary equation

The auxiliary equation for $y''+ay'+by=0$ is $m^2+am+b=0$ . Its roots determine the form of the general solution. The numbers $m$ are exponents, not values of $x$ or $y$ .

If $y_1$ and $y_2$ solve this homogeneous equation, so does $Ay_1+By_2$ : differentiation and multiplication distribute over the sum, so the left-hand side becomes $A(0)+B(0)$ . Distinct exponentials supply two independent choices.

Distinct real roots

For real constants $a$ , $b$ , if $m^2+am+b=0$ has distinct real roots $m_1$ and $m_2$ , then

y=Ae^{m_1x}+Be^{m_2x}

where $A$ and $B$ are independent arbitrary real constants. This case occurs when $a^2-4b>0$ .
Not in the formula booklet: learn it.

Solve the quadratic before writing the function; the sign of each root becomes the sign in its exponent.

Worked example 1: two distinct real roots

Question. Find the general solution of $2y''=2y'+12y$ .

Plan. Rearrange and divide by two, solve the auxiliary equation, then attach one independent constant to each exponential.

\begin{aligned} y''-y'-6y&=0 &&\text{rearrange and divide by }2\\ m^2-m-6&=0 &&\text{auxiliary equation}\\ (m-3)(m+2)&=0 &&\text{factorise}\\ m&=3\text{ or }-2 &&\text{distinct roots}\\ y&=Ae^{3x}+Be^{-2x} &&\text{two exponential solutions} \end{aligned}

Why this step? The roots are rates in the exponentials; $y=A(3)+B(-2)$ would just be a constant, not the required family.

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

Check. Substitution gives coefficients $3^2-3-6=0$ and $(-2)^2-(-2)-6=0$ , so both exponential terms satisfy the equation.

Change only the coefficient of $y$ from $-6$ to $-2$ . Recalculate the roots instead of carrying the old exponents forward.

Worked example 2: changing one coefficient

Question. Find the general solution of $2y''=2y'+4y$ .

Plan. Factorise the new auxiliary quadratic, then use its two roots.

\begin{aligned} y''-y'-2y&=0 &&\text{rearrange and divide by }2\\ m^2-m-2&=0 &&\text{auxiliary equation}\\ (m-2)(m+1)&=0 &&\text{factorise}\\ m&=2\text{ or }-1 &&\text{read the roots}\\ y&=Ae^{2x}+Be^{-x} &&\text{distinct-root form} \end{aligned}

Why this step? The coefficient of $y$ affects the product of the roots and therefore both exponential rates.

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

Check. The two coefficients after substitution are $4-2-2=0$ and $1+1-2=0$ .

An absent $y$ term is allowed: $y''+2y'=0$ has roots $0$ , $-2$ , so its solution is $A+Be^{-2x}$ . The zero root gives the constant function $e^{0x}=1$ .

Watch out

Wrong: $m=-2$ gives $e^{2x}$ . Right: the root itself is the exponent coefficient, so it gives $e^{-2x}$ .

Vary the final coefficient again, keeping track of both signs.

A repeated auxiliary root

When both roots are $m$ , writing $Ae^{mx}+Be^{mx}$ gives only $(A+B)e^{mx}$ . One independent constant has disappeared. The missing second solution is $xe^{mx}$ .

You can see why without guessing. A repeated root means $a=-2m$ , $b=m^2$ . Substitute $y=u(x)e^{mx}$ , where $u$ is an unknown function:

\begin{aligned} y'&=e^{mx}(u'+mu) &&\text{product rule}\\ y''&=e^{mx}(u''+2mu'+m^2u) &&\text{differentiate again}\\ y''-2my'+m^2y&=e^{mx}u'' &&\text{cancel the other terms} \end{aligned}

Thus $u''=0$ , giving $u=A+Bx$ . The extra factor of $x$ restores the second independent constant.

Repeated real root

If $m^2+am+b=0$ has repeated root $m=-a/2$ , then

y=(A+Bx)e^{mx}

where $A$ and $B$ are independent arbitrary real constants. This case occurs when $a^2-4b=0$ .
Not in the formula booklet: learn it.

A repeated root is one root counted twice, so its general solution is structurally different from the distinct-root case.

Worked example 3: recognising a repeated root

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

Plan. Factorise the auxiliary equation and use the repeated-root form.

\begin{aligned} m^2-4m+4&=0 &&\text{auxiliary equation}\\ (m-2)^2&=0 &&\text{perfect square}\\ m&=2\text{ repeated} &&\text{one root counted twice}\\ y&=(A+Bx)e^{2x} &&\text{include the }x\text{ factor} \end{aligned}

Why this step? The functions $e^{2x}$ and $xe^{2x}$ are independent; two copies of $e^{2x}$ are not.

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

Check. The derivation above with $m=2$ reduces the left-hand side to $e^{2x}u''=0$ , since $u=A+Bx$ .

Change only the sign of the coefficient of $y'$ . The root changes sign but remains repeated, so the factor of $x$ must remain.

Worked example 4: a negative repeated root

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

Plan. Solve the auxiliary equation and keep the repeated-root structure.

\begin{aligned} m^2+4m+4&=0 &&\text{auxiliary equation}\\ (m+2)^2&=0 &&\text{factorise}\\ m&=-2\text{ repeated} &&\text{read the sign}\\ y&=(A+Bx)e^{-2x} &&\text{repeated-root form} \end{aligned}

Why this step? A negative exponent changes growth to decay as $x$ increases; it does not remove the second independent solution.

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

Check. Taking $A=0$ , $B=1$ gives $y'=e^{-2x}(1-2x)$ , $y''=e^{-2x}(4x-4)$ . Hence $y''+4y'+4y=e^{-2x}[(4x-4)+4(1-2x)+4x]=0$ .

The repeated-root rule also includes $m=0$ : $y''=0$ gives $y=A+Bx$ , matching integration twice.

Watch out

Wrong: $y=Ae^{-2x}+Be^{-2x}$ for a repeated root. Right: $y=(A+Bx)e^{-2x}$ , with the extra $x$ multiplying one term.

Apply the same structure when the repeated root is a different negative number.

Complex auxiliary roots

A negative discriminant does not mean that a real solution is impossible. Conjugate complex roots combine into real sine and cosine functions.

If the roots are $\alpha\pm i\beta$ , where $\beta>0$ , Euler's formula gives

e^{(\alpha\pm i\beta)x}=e^{\alpha x}(\cos\beta x\pm i\sin\beta x)

Adding the two functions gives $2e^{\alpha x}\cos\beta x$ ; subtracting and dividing by $i$ gives $2e^{\alpha x}\sin\beta x$ . Both are real solutions, so arbitrary real multiples give the required real family. The root's real part controls the exponential factor; the imaginary part controls the trigonometric argument.

Conjugate complex roots

For auxiliary roots $\alpha\pm i\beta$ , with $\alpha$ real and $\beta>0$ , the real general solution is

y=e^{\alpha x}(A\cos\beta x+B\sin\beta x)

where $A$ , $B$ are independent arbitrary real constants. This case occurs when $a^2-4b<0$ .
Not in the formula booklet: learn it.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Core

Statistics

Mechanics

Discrete Mathematics

Additional Pure Mathematics

4.10d - Second order homogeneous equations

The auxiliary equation and distinct roots

A repeated auxiliary root

Complex auxiliary roots

Continue this lesson.

Learn faster with Nova