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 \(y_1=e^{3 t}\) and \(y_2=e^{-t}\) are solutions of the ODE \[\frac{\dd^2y}{\dd t^2}-2\frac{\dd y}{\dd t}-3 y(t)=0\]

We have \[y_1=e^{3 t},\qquad \frac{\dd y_1}{\dd t}= 3 e^{3 t},\qquad \frac{\dd^2y_1}{\dd t^2}=9 e^{3 t}\] So: \[\frac{\dd^2y_1}{\dd t^2}-2\frac{\dd y_1}{\dd t}-3 y_1(t)= 9 e^{3 t} -2\times 3e^{3 t}-3 e^{3 t} = e^{3 t}(9-6-3)=0\]
Likewise we have \[y_2=e^{-t},\qquad \frac{\dd y_2}{dt}= - e^{-t},\qquad \frac{\dd^2y_2}{\dd t^2}=e^{-t}\] So \[\frac{\dd^2y_2}{\dd t^2}-2\frac{\dd y_2}{\dd t}-3 y_2(t)= e^{-t} -2\times -e^{-t}-3 e^{-t} = e^{-t}(1+2-3)=0\]

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.