Further Maths/Lessons/Differential Equations/Homogeneous Equations
Core Pure · Methods in Differential Equations · Year 2

Second-Order Homogeneous Differential Equations

The auxiliary equation and the three cases for its roots: real distinct, repeated, and complex.

9 min read 2 worked examples

Second-order homogeneous equations

Definition 7.3 — the auxiliary equation

A second-order linear homogeneous differential equation with constant coefficients has the form

$$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0$$

Trying a solution $y=e^{mx}$ gives $\dfrac{dy}{dx}=me^{mx}$ and $\dfrac{d^2y}{dx^2}=m^2e^{mx}$. Substituting and dividing by $e^{mx}$ (never zero) gives the auxiliary equation:

$$am^2+bm+c=0$$

Theorem 7.4 — general solution by case

Let $m_1,m_2$ be the roots of the auxiliary equation.

Two distinct real roots $m_1\neq m_2$: $\;y=Ae^{m_1x}+Be^{m_2x}$.

Repeated real root $m_1=m_2=m$: $\;y=(A+Bx)e^{mx}$.

Complex conjugate roots $m=p\pm qi$: $\;y=e^{px}(A\cos qx+B\sin qx)$.

ProofThe repeated-root case needs an extra factor of x
1
If $m_1=m_2=m$, then $y=e^{mx}$ alone only supplies one arbitrary constant, but a second-order equation needs two independent solutions.
2
Try $y=xe^{mx}$ as the second solution. Then $\dfrac{dy}{dx}=e^{mx}+mxe^{mx}$ and $\dfrac{d^2y}{dx^2}=2me^{mx}+m^2xe^{mx}$.
3
Substitute into $a\dfrac{d^2y}{dx^2}+b\dfrac{dy}{dx}+cy$: $a(2me^{mx}+m^2xe^{mx})+b(e^{mx}+mxe^{mx})+cxe^{mx}$.
4
Group: $e^{mx}(2am+b)+xe^{mx}(am^2+bm+c)$. The second bracket is zero since $m$ satisfies the auxiliary equation; the first bracket is zero because a repeated root of $am^2+bm+c=0$ satisfies $m=-\dfrac b{2a}$, i.e. $2am+b=0$. So $y=xe^{mx}$ is indeed a solution, and by linearity $y=(A+Bx)e^{mx}$ is the general solution.
∎
Example 1 · Distinct real roots
Find the general solution to $\dfrac{d^2y}{dx^2}+3\dfrac{dy}{dx}-10y=0$.

Auxiliary equation: $m^2+3m-10=0 \Rightarrow (m+5)(m-2)=0 \Rightarrow m=-5,2$.

General solution: $y=Ae^{-5x}+Be^{2x}$.

Example 2 · Complex roots
Find the general solution to $\dfrac{d^2y}{dx^2}-6\dfrac{dy}{dx}+25y=0$.

Auxiliary equation: $m^2-6m+25=0 \Rightarrow m=\dfrac{6\pm\sqrt{36-100}}2=3\pm4i$.

General solution: $y=e^{3x}(A\cos4x+B\sin4x)$.

Exam-style questions

Question 1 4 marks

Find the general solution to $\dfrac{d^2y}{dx^2}+8\dfrac{dy}{dx}+16y=0$.

$m^2+8m+16=0\Rightarrow(m+4)^2=0\Rightarrow m=-4$ (repeated). (M1, A1)

$y=(A+Bx)e^{-4x}$. (A1, A1)

Question 2 5 marks

Find the general solution to $\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+5y=0$.

Auxiliary equation: $m^2-2m+5=0$. (M1)

$m=\dfrac{2\pm\sqrt{4-20}}2=\dfrac{2\pm4i}2=1\pm2i$. (M1, A1)

Since the roots are complex, $p=1,q=2$, so the general solution is $y=e^x(A\cos2x+B\sin2x)$. (A1, A1)

← Previous
First-Order Equations