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? $\endgroup$ – Arturo don Juan Jun 24 '14 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 '14 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 '14 at 16:44
  • $\begingroup$ @whosleon Why don't you add that as an answer? $\endgroup$ – M. Vinay Jun 27 '14 at 17:02

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.


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


the whosleon's answer is pretty good but I just wanna add something. the particular solution is outside of homogeneous solution space, that means you cant write yp = c1y1 + c2y2. so you have to jump into another dimension by chaning either y1 or y2's dimension using u1 and u2. you could call u1 and u2 the dimension booster. (for example, consider x+y=0 the homogeneous solution is a line pass the (0,0) but if x+y = c , the line will move up and down. you can never jump when you only moving along the line. so you need boost the dimension (again since the homogeneous equal 0, it is impossible to get something other than 0 within solution space)) since y is function of t(or x or whatever you call it) and as well as u1 and u2. you can achieve any dimension you want. so yp can be expressed as yp = u1y1 + u2y2 same to the prime of everything. since the second order differential equation need y and y' to determine the whole thing(its like if you know postion and velocity of a object attached by spring you will determine the whole process) so y and y' are independent basis. it is nonsense to say you have to have 5 m/s when you at the origin of the spring, you can have whatever speed you want thats why they are independent. Since y and y' are independent and form the basis of the solution, the differential equation doesn't care product rule anymore, thats why you can write yp = u1y1 + u2y2 yp'= u1y1' + u2y2' again the reason y and y' use the same u is they form a basis. but in reality there is a product rule (ab)' = a'b + ab' the only way to obtain above is to set any u' = 0 to achieve the mapping process. so ui'yi = 0 I know its kinda painful to understand, the main point is to have independent basis combination you have to get ride of those term. maybe the spring example will help alot. if you know the physics, homogeneous mean no external force where non homo mean there is external force. when you have external force, you need a "dimension booster" to get to the "process" that original physics process can never achieve. im a physics guy so my explanation is bit physical but I think that will cover your question. Sorry for the formatting, im lazy.

  • 1
    $\begingroup$ Please don't be lazy - do format your post! $\endgroup$ – Mårten W Jun 29 '16 at 21:43

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