Characteristic equation

The characteristic equation of a constant-coefficient linear ODE is the polynomial equation you get by substituting the trial solution $y=e^{rx}$ and dividing through. Its roots determine the general solution.

For a second-order ODE $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$ (constants $a,b,c$), the trial $y=e^{rx}$ gives $y^{\prime }=r{e}^{rx}$, $y^{\prime \prime }=r^2{e}^{rx}$. Substituting:

$e^{rx}(a{r}^2+br+c)=0$

Since $e^{rx}\ne 0$, we need

$\boxed{a{r}^2+br+c=0}$

This is the characteristic equation. Its roots are the values of $r$ that make $e^{rx}$ a solution.

Why $e^{rx}$

A second-order ODE relates $y,y^{\prime },y^{\prime \prime }$. For $y=e^{rx}$, all three are proportional to $e^{rx}$ — the only difference is the coefficient ($1,r,r^2$). So the equation $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$ becomes $e^{rx}(a{r}^2+br+c)=0$, a polynomial in $r$.

Polynomials we can solve. ODEs we generally can’t.

Characteristic equation in LTI systems

The same polynomial shows up under a different name in linear time-invariant systems. Take the Transfer function $H(s)=N(s)/D(s)$ of an LTI system. Setting the denominator to zero,

$D(s)=0$

is the characteristic equation of the system, and its roots are the poles of $H(s)$.

The connection to ODEs is direct. For a constant-coefficient ODE $a{y}^{\prime \prime }+b{y}^{\prime }+cy=g(t)$ with zero initial conditions, Laplace-transforming gives $(a{s}^2+bs+c)Y(s)=G(s)$, so $H(s)=1/(a{s}^2+bs+c)$. The denominator $a{s}^2+bs+c$ is exactly the characteristic polynomial you’d get from the trial $y=e^{rx}$. Same polynomial, two viewpoints: $r$ in the time-domain trial vs. $s$ in the Laplace-domain transfer function.

So a root $r_i$ of the characteristic equation is a pole $s_i$ of the transfer function, and the corresponding natural mode of the system is $e^{r_{i}t}$ — exactly the homogeneous solution. Pole locations determine stability and transient behavior: left-half-plane poles ($\mathrm{Re}(s)<0$) give decaying modes, right-half-plane poles give growing (unstable) modes, and imaginary-axis poles give sustained oscillation.

Three cases based on the discriminant

The discriminant $\Delta =b^2-4ac$ tells you the nature of the roots:

Case 1: $\Delta >0$ — distinct real roots

Two real roots $r_1\ne r_2$. Two linearly independent solutions:

$y_1(x)=e^{r_{1}x},y_2(x)=e^{r_{2}x}$

Wronskian: $W=e^{r_{1}x}\cdot r_2{e}^{r_{2}x}-r_1{e}^{r_{1}x}\cdot e^{r_{2}x}=(r_2-r_1)e^{(r_1+r_2)x}\ne 0$.

General solution:

$y(x)=c_1{e}^{r_{1}x}+c_2{e}^{r_{2}x}$

Example: $2y^{\prime \prime }+7y^{\prime }-4y=0$. Characteristic: $2r^2+7r-4=0$, roots $r=-4,\frac{1}{2}$. Solution: $y=c_1{e}^{-4x}+c_2{e}^{x/2}$.

Case 2: $\Delta =0$ — repeated real root

One root $r_1=r_2=-\frac{b}{2a}$. The first solution is $y_1=e^{r_{1}x}$.

(Note: the Differential Equations lecture PDF stops here for this case, leaving the second-solution derivation incomplete. The actual result, derived below, is $y_2(x)=x{e}^{r_{1}x}$.)

To get a second linearly independent solution, multiply by $x$:

$y_2(x)=x{e}^{r_{1}x}$

Direct verification: $y_2^{\prime }=e^{r_{1}x}(1+r_{1}x)$ and $y_2^{\prime \prime }=e^{r_{1}x}(2r_1+r_1^{2}x)$. Substituting into $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$:

$e^{r_{1}x}\left[(2a{r}_1+b)+x(a{r}_1^2+b{r}_1+c)\right]$

The second bracket is the characteristic equation evaluated at $r_1$, which is zero. The first bracket is $2a{r}_1+b=2a\cdot (-b/(2a))+b=-b+b=0$ (using the repeated-root condition $r_1=-b/(2a)$). So the whole expression vanishes — $y_2$ is indeed a solution. Independent of $y_1$ because they’re not proportional.

(Or derive systematically via Method of reduction of order, starting from $y_1$.)

General solution:

$y(x)=(c_1+c_{2}x)e^{r_{1}x}$

Case 3: $\Delta <0$ — complex conjugate roots

Two complex roots $r=\alpha \pm i\beta$ where $\alpha =-\frac{b}{2a}$, $\beta =\frac{\sqrt{4ac-b^2}}{2a}$.

The complex-valued solutions $e^{(\alpha +i\beta )x}$ and $e^{(\alpha -i\beta )x}$ aren’t useful for real-valued problems. By Euler’s formula:

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

For an ODE with real coefficients, if $\varphi =u+iv$ is a complex-valued solution, then $u(x)=\mathrm{Re}(\varphi )$ and $v(x)=\mathrm{Im}(\varphi )$ are real-valued solutions of the same equation.

Taking real and imaginary parts:

$Y_1(x)=e^{\alpha x}\cos (\beta x),Y_2(x)=e^{\alpha x}\sin (\beta x)$

These are linearly independent (Wronskian = $\beta e^{2\alpha x}\ne 0$). General solution:

$y(x)=e^{\alpha x}(c_1\cos (\beta x)+c_2\sin (\beta x))$

Example: $y^{\prime \prime }-y^{\prime }+y=0$, $y(0)=1$, $y^{\prime }(0)=-2$.

Characteristic: $r^2-r+1=0$, roots $r=\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$. So $\alpha =1/2$, $\beta =\sqrt{3}/2$.

General solution: $y(x)=e^{x/2}\left[c_1\cos \left(\frac{\sqrt{3}}{2}x\right)+c_2\sin \left(\frac{\sqrt{3}}{2}x\right)\right]$.

Apply initial conditions to find $c_1,c_2$ — straightforward but tedious.

Higher-order

For $n$-th order ODEs, the characteristic polynomial is degree $n$. For each root $r$ of multiplicity $m$, the contribution to the general solution is $(c_0+c_{1}x+c_2{x}^2+\cdots +c_{m-1}{x}^{m-1})e^{rx}$.

For example, $r=2$ with multiplicity 3 contributes $(c_0+c_{1}x+c_2{x}^2)e^{2x}$ — three linearly independent solutions.

For complex roots in higher orders, you pair conjugates and use the real form $e^{\alpha x}[\cos (\beta x),\sin (\beta x)]$ analogously.

In context

The characteristic equation handles all constant-coefficient homogeneous linear ODEs. For variable coefficients, you need Method of reduction of order or, for second-order, sometimes power series methods (not in this vault). For nonhomogeneous equations, find the homogeneous part with the characteristic equation, then add a particular solution via Method of undetermined coefficients or Method of variation of parameters.