Differential Eequations: Second Order Linear with Constant Coefficients

In this subsection, we look at equations of the form $$ a\,\frac{d^2 y}{dx^2}+b\,\frac{dy}{dx}+c\,y=f(x), $$ where a, b and c are constants. We start with the case where f(x)=0, which is said to be {\bf homogeneous in y}. We'll need the following key fact about linear homogeneous ODEs.

Theorem 1.1

If f(x) and g(x) are both solutions of a linear homogeneous ODE, then so is $$A\,f(x)+B\,g(x),$$ where A and B are constants.

We'll also need to use the fact that the general solution of a second-order ODE always has two unknown constants.

Let's focus on the following example: $$ \frac{d^2 y}{dx^2}-4\,\frac{dy}{dx}+3\,y=0. $$ We reason as follows. In this equation, by adding a multiple of y to multiples of its first and second derivatives, we get zero. So a good guess is that y is an exponential function: the only class of function that, when differentiated, gives a multiple of itself. We therefore look for a solution of the form $$ y=e^{\lambda x}, $$ where \lambda is a constant. If we make this assumption, we obtain, by differentiating twice,

\begin{eqnarray} \frac{dy}{dx}&=&\lambda\,e^{\lambda x},\\ \frac{d^2y}{dx^2}&=&\lambda^2\,e^{\lambda x}. \end{eqnarray}

Substituting into our original ODE gives $$ (\lambda^2-4\,\lambda+3)\,e^{\lambda x}=0. $$ Cancelling, we obtain $$ \lambda^2-4\,\lambda+3=0, $$ whose solutions are \lambda=1 and \lambda=3.

So it looks as if e^x and e^{3x} are both solutions of the differential equation.

But by Theorem 1.1, we therefore know that $$ y=A\,e^x+B\,e^{3x} $$ is also a solution, where A and B are any constants. Since this is a solution containing two unknown constants, it must be the general solution of the differential equation.

If the roots of the auxiliary equation are non-real and conjugate, of the form \lambda=p\pm q\,i, then by the above reasoning the general solution ought to be $$ y=A\,e^{(p+q\,i)\,x}+B\,e^{(p+q\,i)\,x}. $$ This looks decidedly odd, since we would expect the solution to be real. We can get round this problem by recalling that $$e^{i\,q\,x}=\cos(qx)+i\,\sin(qx),$$ and rewriting the above as $$y=e^{p\,x}\,[P\,\cos(qx)+Q\,\sin(qx)],$$ where P=\frac{1}{2}(A+B) and Q=\frac{1}{2}i(A-B) are real constants.

The case we've yet to treat is that where the roots are real but repeated. In this case, it's easy to show that both e^{\lambda x} and x\,e^{\lambda x} satisfy the differential equation, and therefore that the general solution is $$y=e^{\lambda x}\,(A+B\,x).$$