I have yet to find a decent answer on this, and so I don't think this question is inappropriate. Also, this question is mainly meant for people that are very familiar with this method.

In the method of variation of parameters method for solving the non-homogeneous (heterogeneous?) part of a second order linear differential equation - that is, one in the form $$y''(t)+a(t)y'(t)+b(t)y(t)=g(t)$$I'm still boggled with the exact purpose of setting parts of the first derivative of the particular solution equal to solution.

I will explain what I'm talking about here.

Suppose we have a $2^{nd}$ order linear non-homogeneous ODE, $$y''(t)+a(t)y'(t)+b(t)y(t)=g(t)$$ Also, suppose we've already obtained a general solution for the homogeneous version of the ODE, $y''(t)+a(t)y'(t)+b(t)y(t)=0$, that's in the nice form $$y_h=c_1y_1+c_2y_2$$ To get the particular solution for the non-homogeneous part, we start off with the weird (but successful) method of variation of parameters - change the $c_1$ and $c_2$ to $u_1(t)$ and $u_2(t)$ respectively (hence, variation of parameters), and call that $y_p$.


The goal is to find out $u_1(t)$ and $u_2(t)$, and so we now get the $1^{st}$ and $2^{nd}$ derivatives of $y_p$ so we can slap them into the original ODE and hopefully solve for $u_1(t)$ and $u_2(t)$. The first derivative follows by chain rule:

$$y'_p = u'_1(t)y_1 + u_1(t)y'_1 + u'_2(t)y_2 + u_2(t)y'_2$$

Here is the part I don't understand.

In every textbook I've seen this in, the next step goes basically as follows:

For no reason and no motivation (as far as we're concerned right now), we now set the following: $$u'_1(t)y_1 + u'_2(t)y_2 = 0$$ This will turn out to work in the end very nicely, and it will also make our computations much more simple.

I've seen full well that this does work out nicely, and that it does make the computations much more simple, but my question is why is this done, and why does it work out so nicely? Why does that equation $$u'_1(t)y_1 + u'_2(t)y_2 = 0$$ always have to be satisfied for particular solutions to $2^{nd}$ order linear non-homogeneous ODE's of this form?.

  • $\begingroup$ Could somebody answer this please? $\endgroup$ Jun 24, 2014 at 17:46
  • $\begingroup$ The explanation that you are looking for uses the fact that $y_1$ and $y_2$, being linearly independent solutions to the homogeneous equation, form a basis for the space of solutions of the inhomogeneous equation. This means that the solution vector $\begin{pmatrix} y_p \\ y_p^{\prime} \end{pmatrix}$ can be written as a linear combination of the vectors $\begin{pmatrix} y_1 \\ y_1^{\prime} \end{pmatrix}$ and $\begin{pmatrix} y_2 \\ y_2^{\prime} \end{pmatrix}$. $\endgroup$
    – whosleon
    Jun 27, 2014 at 16:44
  • $\begingroup$ Namely, that we can write: $\begin{pmatrix} y_p \\ y_p^{\prime} \end{pmatrix} = u_1\begin{pmatrix} y_1 \\ y_1^{\prime} \end{pmatrix} + u_2\begin{pmatrix} y_2 \\ y_2^{\prime} \end{pmatrix}$ By considering the two components we get two equations for $y_p$, and differentiating the first and equating it with the second tells us that $u_1^{\prime}y_1 + u_2^{\prime}y_2 = 0$ $\endgroup$
    – whosleon
    Jun 27, 2014 at 16:44
  • $\begingroup$ @whosleon Why don't you add that as an answer? $\endgroup$
    – M. Vinay
    Jun 27, 2014 at 17:02

2 Answers 2


The explanation that you are looking for relies on the fact that $y_1$ and $y_2$, being linearly independent solutions to the homogeneous equation, form a basis for the solution space of the inhomogeneous equation. This means that they span the space, so that the solution vector $\begin{pmatrix}y_p \\ y_p^{\prime}\end{pmatrix}$ can be written as a linear combination of the vectors $\begin{pmatrix}y_1 \\ y_1^{\prime}\end{pmatrix}$ and $\begin{pmatrix}y_2 \\ y_2^{\prime}\end{pmatrix}$. Hence, for some functions $u_1(t)$ and $u_2(t)$: $$\begin{pmatrix}y_p \\ y_p^{\prime}\end{pmatrix} = u_1\begin{pmatrix}y_1 \\ y_1^{\prime}\end{pmatrix} + u_2\begin{pmatrix}y_2 \\ y_2^{\prime}\end{pmatrix}$$ We consider the two components: $$y_p = u_1y_1 + u_2y_2$$ $$y_p^{\prime} = u_1y_1^{\prime} + u_2y_2^{\prime}$$ By differentiating the first equation and subtracting it from the second we get that $$0 = u_1^{\prime}y_1 + u_2^{\prime}y_2$$ which is what you were looking for. In my opinion, the subtlety lies in the observation that the above vectors form a basis for the solution space. Perhaps someone has a good explanation of why that is the case.

  • 2
    $\begingroup$ This is not a good answer. $\endgroup$ Feb 12, 2020 at 16:43
  • 6
    $\begingroup$ Two linearly independent solutions to the homogeneous equation absolutely do not form a basis for the solution space to the inhomogeneous equation. Your answer is just wildly false. Of course we can play a game of osculating parameters and magically it works out, but it is not motivated at all. The method is essentially analytic, it requires an analysis of the Green's functions for the equation and an analytic approach to piecing them together via something like distribution theory, it is not just linear algebra (although, obviously, with the theorem in hand the computations are just algebra). $\endgroup$ Feb 12, 2020 at 16:49
  • $\begingroup$ @MarcelBesixdouze Glad I read your comment, and you are right... what we can say at the very least is that the solutions of the inhomogeneous equation forms the subspace $u + V$ where $u$ is a solution of the inhom. and $V$ is the space of the solutions of the homogeneous equation (Amann ODE book thrm. 11.12). This is really a lesson for me: every time I've seen an upvoted accepted question, I've assumed it says the undisputed mathematical truth, but this time it's just plain wrong and misleading. $\endgroup$
    – César VB
    Oct 26, 2023 at 8:49

This is an extended comment on @whosleon answer.

I want to point out that to write $\begin{pmatrix}y_p(t) \\ y_p^{\prime}(t)\end{pmatrix} = u_1(t)\begin{pmatrix}w_1(t) \\ z_1(t)\end{pmatrix} + u_2(t)\begin{pmatrix}w_2(t) \\ z_2(t)\end{pmatrix}$ with coefficients being functions $u_1(t)$ and $u_2(t)$ we just need to have two vectors $\begin{pmatrix}w_1(t) \\ z_1(t)\end{pmatrix}$ and $\begin{pmatrix}w_2(t) \\ z_2(t)\end{pmatrix}$ be linearly independent for each $t$. This is true for $\begin{pmatrix}w_1(t) \\ z_1(t)\end{pmatrix}=\begin{pmatrix}y_1(t) \\ y_1^{\prime}(t)\end{pmatrix}$, $\begin{pmatrix}w_2(t) \\ z_2(t)\end{pmatrix}=\begin{pmatrix}y_2(t) \\ y_2^{\prime}(t)\end{pmatrix}$ where $y_1(t)$ and $y_2(t)$ are independent solutions to any homogeneous linear second order ODE, no matter what specific equation we consider; it is also true for $\begin{pmatrix}w_1(t) \\ z_1(t)\end{pmatrix}=\begin{pmatrix} 1\\0\end{pmatrix}, \begin{pmatrix}w_2(t) \\ z_2(t)\end{pmatrix}=\begin{pmatrix} 0\\1\end{pmatrix}$.

This does not explain why using the solutions of the associated homogeneous equation $\begin{pmatrix}w_1(t) \\ z_1(t)\end{pmatrix}=\begin{pmatrix}y_1(t) \\ y_1^{\prime}(t)\end{pmatrix}$, $\begin{pmatrix}w_2(t) \\ z_2(t)\end{pmatrix}=\begin{pmatrix}y_2(t) \\ y_2^{\prime}(t)\end{pmatrix}$ is a good choice, in the sense of giving computable $u_1(t)$ and $u_2(t)$.

My attempt at writing one possible explanation is in the answer to Intuition behind variation of parameters method for solving differential equations

Another potentially helpful idea is in an answer here: second order ODE via variation of parameters


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