Take a non-homogeneous second order ode with constant coefficients $$ y''(t)+py'(t)+qy(t)=f(t),$$ the solutions of this equation have the following form $$ t\mapsto y_{\text{part}}(t)+A y_1(t)+By_2(t),$$ where $y_{\text{part}}$ is a particular solution of this non-homogeneous second order ode and $y_1$ and $y_2$ are independent solutions of the associated homogeneous second order ode. The question is: how to determine $y_{\text{part}}$?

The first option is the variation of parameters: one can choose to find $y_{\text{part}}$ with the following form $t\mapsto A(t) y_1(t)+B(t)y_2(t),$ where $A$ and $B$ satisfy the following equation $$ A'(t) y_1(t)+B'(t)y_2(t)=0, \tag{1} $$ and $$ A'(t) y_1'(t)+B'(t)y_2'(t)=f(t). \tag{2} $$ If one knows explicitly $y_1$ and $y_2$, one could find $A$ and $B$ by integration.

The second option is to put this second order ode in the framework of first order ode. One can set $$Y(t)=\begin{pmatrix} y(t) \\ y'(t) \end{pmatrix},$$ then $Y$ satisfies the following first order ode: $$Y'(t) = \begin{pmatrix} 0 & 1 \\ q & p \end{pmatrix} Y(t) + \begin{pmatrix} 0 \\ -f(t) \end{pmatrix}.$$ Then the general solution of this ode can be written as $$ Y(t) = \exp\left( \begin{pmatrix} 0 & 1 \\ q & p \end{pmatrix} t\right) Y_0 + \int_0^t \exp\left( \begin{pmatrix} 0 & 1 \\ -q & -p \end{pmatrix} (t-s)\right) \begin{pmatrix} 0 \\ -f(s) \end{pmatrix}\text{d}s$$

My question is: how are these two options related? In particular, can we explain where does the hypothesis $(1)$ come from with the second option?


1 Answer 1


Regarding the first option, I was told that condtion (1) is only assumed for the sake of simplicity. It's an extra condition, in no way related to (2). Remember that you only need to find one particular solution.

In the second option, you are simply transforming your second order ODE into a first order differential equations system. Note that the general solution of Y(t) is very similar to the general solution of a first order ODE.

I don't think they are related. At least, nothing comes to my mind. For equations of higher order, however, the second method is, supposedly, easier.

  • $\begingroup$ Have you ever seen the method of variation of parameters for ode of order $\ge 3$? $\endgroup$
    – user37238
    Commented Jan 18, 2017 at 8:52
  • 1
    $\begingroup$ @user37238 It exists and it is described in the textbooks. The key trick of doing variation of parameters is to obtain a linear system with Wronski matrix (which is always non-degenerate if you've picked basis solutions). $\endgroup$
    – Evgeny
    Commented Jan 18, 2017 at 8:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .