Homogeneous linear ODEs of order 2 with constant coefficients

Ordinary differential equations: Second-order linear ODEs with constant coefficients

Homogeneous linear ODEs of order 2 with constant coefficients

A differential equation of the form

$a\frac{\dd^2y}{\dd t^2}+b\frac{\dd y}{\dd t}+c\,y=0$ where $a$ , $b$ and $c$ are real constants with $a\neq 0$ is called a homogeneous linear second-order differential equation with constant coefficients.

The word means linear in this respect two things:

If $y(t)$ is a solution of the differential equation, then $A\cdot y(t)$ is also a solution.
If $y_1(t)$ and $y_2(t)$ are solutions of the differential equation, then $y_1(t)+y_2(t)$ is also a solution.

You can combine these two statements by saying that for any two solutions $y_1(t)$ and $y_2(t)$ of the differential equation also any linear combination $A\,y_1(t)+B\,y_2(t)$ is a solution.

Show that $y_1=e^{-t}$ and $y_2=e^{3 t}$ are solutions of the ODE

$\frac{\dd^2y}{\dd t^2}-2\frac{\dd y}{\dd t}-3 y(t)=0$

We have

$y_1=e^{-t},\qquad \frac{\dd y_1}{\dd t}= - e^{-t},\qquad \frac{\dd^2y_1}{\dd t^2}= e^{-t}$ So:

$\frac{\dd^2y_1}{\dd t^2}-2\frac{\dd y_1}{\dd t}-3 y_1(t)= e^{-t} -2\times -e^{-t}-3 e^{-t} = e^{-t}(1+2-3)=0$
Likewise we have

$y_2=e^{3 t},\qquad \frac{\dd y_2}{dt}= 3 e^{3 t},\qquad \frac{\dd^2y_2}{\dd t^2}=9e^{3 t}$ So

$\frac{\dd^2y_2}{\dd t^2}-2\frac{\dd y_2}{\dd t}-3 y_2(t)= 9 e^{3 t} -2\times 3e^{3 t}-3 e^{3 t} = e^{3 t}(9-6-3)=0$

New example

Characteristic equation The general solution of the differential equation of the form

$a\frac{\dd^2y}{\dd t^2}+b\frac{\dd y}{\dd t}+c\,y=0$ we get trying exponential fucntions as solutions. Suppose that

$y(t)=e^{\lambda t}$ is a solution, then substitution in the ODE gives the equation

$a\,\lambda^2e^{\lambda t}+b\,\lambda e^{\lambda t}+c\, e^{\lambda t}=0$ In other words, after dividing by $e^{\lambda t}$ ,

$a\,\lambda^2+b\,\lambda + c=0$ This is called the characteristic equation of the differential equation. Every root of this quadratic equation gives a solution $e^{\lambda t}$ .

The nature of the solutions is determined by the sign of the discriminant $D=b^2-4ac$ . If $D>0$ , we have two real roots. If $D=0$ we have one real root. Finally, if $D<0$ , we have complex roots. We will discuss the solutions of the differential equation for each of the three cases.