A 2nd order linear equation that can be written in the form \(ay''+by'+cy=f\) where \(a,b,c\) are constants, \(a\neq 0\), and \(f\) is a function (of \(x\) or \(t\)) is called constant coefficient.

Consider a 2nd order linear homogeneous constant coefficient equation: \(ay''+by'+cy=0\).

We guess that there might be solutions of the form \(y=e^{rt}\); so \(y'=re^{rt}\) and \(y''=r^2 e^{rt}\).

Then \(ay''+by'+cy=0\) becomes \(ar^2 e^{rt}+br e^{rt}+c e^{rt}=0\), equivalently \(\left (ar^2+br +c\right ) e^{rt}=0\).

Since \(e^{rt}\) is never \(0\), \(ar^2+br +c=0\). This is called the characteristic equation and the left side is called the characteristic polynomial.

This equation can (1) have two distinct real solutions \(r_1<r_2\), or (2) have just one real solutions \(r_1\), or (3) have complex conjugate solutions \(\lambda \pm \omega i\) where \(\lambda ,\omega \) are real numbers, \(\omega \neq 0\), and \(i=\sqrt {-1}\).


Case 1: Assume \(ar^2+br +c=0\) has two distinct real solutions \(r_1<r_2\). Then \(y_1=e^{r_1 t}\) and \(y_2=e^{r_2 t}\) form a fundamental set of solutions of \(ay''+by'+cy=0\) since \(y_2/y_1=e^{(r_2-r_1)t}\) is non-constant (since the derivative of \(y_2/y_1\) is \(W/y_1^2\) and is non-zero, the Wronskian \(W\) isn’t \(0\)).

So the general solution in this case is \(y=c_1 e^{r_1 t}+c_2e^{r_2 t}\) where \(c_1,c_2\) are arbitrary constants.

Case 2: Assume \(ar^2+br +c=0\) has just one real solutions \(r_1\). That means that
\(ar^2+br +c=a(r-r_1)^2=a(r^2-2r_1 r+r_1^2)\) which in particular implies that \(b=-2ar_1\) and \(c=ar_1^2\).

So \(y_1=e^{r_1 t}\) is a solution of \(ay''+by'+cy=0\), but does not form a fundamental set of solutions. Let’s see if we can find another solution of the form \(y_2=uy_1=ue^{r_1 t}\) where \(u\) is a function of \(t\).

\(y_2'=u' y_1+uy_1'\) and \(y''=u''y_1+u'y_1'+u'y_1'+uy_1''=u''y_1+2u'y_1'+uy_1''\).

Then

\begin{eqnarray*} ay''+by'+cy=0 &\rightarrow & a(u''y_1+2u'y_1'+uy_1'')+b(u' y_1+uy_1')+c(uy_1)=0 \\ &\rightarrow & au''y_1+2au'y_1'+auy_1''+bu' y_1+buy_1'+cuy_1=0 \\ &\rightarrow & au''y_1+2au'y_1'+bu' y_1+u\left (ay_1''+by_1'+cy_1\right )=0 \\ &\rightarrow & au''y_1+2au'y_1'+bu' y_1=0 \\ &\rightarrow & au''e^{r_1 t}+2au' r_1e^{r_1 t}+bu' e^{r_1 t}=0 \\ &\rightarrow & au''+u'(2ar_1+b) =0 \\ &\rightarrow & au''=0\mbox { since }b=-2ar_1.\\ &\rightarrow & u=m t+k\mbox { by integrating both sides twice}\\ \end{eqnarray*}

Since we only need one solution that is linearly independent with \(y_1=e^{r_1 t}\) we choose \(u=t\), meaning \(y_2=te^{r_1 t}\) as then \(y_2/y_1=t\) is non-constant (and so the Wronskian is non-zero).

So the general solution in this case can be expressed as \(y=(c_1+c_2 t)e^{r_1 t}\) where \(c_1,c_2\) are arbitrary constants.

Case 3: Assume \(ar^2+br +c=0\) has complex conjugate solutions \(\lambda \pm \omega i\) where \(\lambda ,\omega \) are real numbers, \(\omega \neq 0\), and \(i=\sqrt {-1}\).

Then \(\bar {y_1}=e^{(\lambda + \omega i)t}\) and \(\bar {y_2}=e^{(\lambda - \omega i)t}\) are solutions of \(ay''+by'+cy=0\). While any linear combination of these is a solution, some are not real-valued, even for real values of \(t\). We restrict our attention towards finding the general solution for real-valued functions.

\(e^{(\lambda \pm \omega i)t}=e^{\lambda t}e^{\pm \omega i t}=e^{\lambda t}(\cos {(\omega t)}\pm i\sin {(\omega t)})\) using Euler’s formula
\(e^{i\theta }=\cos {(\theta )}+i\sin {(\theta )}\).

So \(\bar {y_1}=e^{\lambda t}(\cos {(\omega t)}+ i\sin {(\omega t)})\) and \(\bar {y_2}=e^{\lambda t}(\cos {(\omega t)}- i\sin {(\omega t)})\).

In particular, any linear combination of these are solutions, so we can take

\[y_{1}=\frac {1}{2}[\bar {y_1}+\bar {y_2}]=\frac {1}{2}\left [e^{\lambda t}(\cos {(\omega t)}+ i\sin {(\omega t)})+e^{\lambda t}(\cos {(\omega t)}- i\sin {(\omega t)})\right ]=e^{\lambda t}\cos {(\omega t)}\mbox { and }\]
\[y_{2}=\frac {1}{2i}[\bar {y_1}-\bar {y_2}]=\frac {1}{2i}\left [e^{\lambda t}(\cos {(\omega t)}+ i\sin {(\omega t)})-e^{\lambda t}(\cos {(\omega t)}- i\sin {(\omega t)})\right ]=e^{\lambda t}\sin {(\omega t)}.\]

Since \(y_{2}/y_{1}=\tan {(\omega t)}\) is non-constant \(y_{1},y_{2}\) form a fundamental set of solutions to
\(ay''+by'+cy=0\), and hence the general solution is \(y=e^{\lambda t}\left [c_{1}\cos {(\omega t)}+c_{2}\sin {(\omega t)}\right ]\) where \(c_1,c_2\) are arbitrary constants.


Examples:

(1) Let’s solve the IVP \(y''=4y'-4y\), \(y(0)=1\), \(y'(0)=0\).

First rewrite the de in standard form, \(y''-4y'+4y=0\). The characteristic equation is \(r^2-4r+4=0\). Since \(r^2-4r+4=(r-2)^2\) the only solution is \(r=2\). So the general solution is \(y=(c_1+c_2 t)e^{2t}\). Then \(y'=c_2 e^{2t}+2(c_1+c_2 t)e^{2t}\).

Since \(y(0)=1\) we get \(1=c_1\). Since \(y'(0)=0\) we get \(0=c_2+2c_1\). Since \(c_1=1\) it follows that \(c_2=-2\). So the IVP solution is \(y=(1-2t)e^{2t}\).

(2) Let’s find the general solution of \(y''+2y'+10y=0\). The characteristic equation is \(r^2+2r+10=0\) and by completing the square it can be rewritten as \((r+1)^2+9=0\) so \(r=-1\pm 3i\). Therefore the general solution is \(y=e^{-t}\left [c_1 \cos {(3t)}+c_2 \sin {(3t)}\right ]\).