# "Variation of Constant" -method to solve linear DYs?

My school instructs to use some method called "variation of constant" (first page here) to solve linear DY more in my earlier question here. I think I solved the problem without the method just by the integrating factor -multiplication. Could someone explain why I get the same solution without using the method to the problem in my earlier question here?

In the foreign course book, the method is very fast covered on the page 636. It mixes up the integrating-factor -method and the variation -method making it for me very hard reading, cannot actually understand a word from it. Then in a rush-minute, it mentions some basic rule "basic rule from analysis" -- and well I think it has a mistake on page 637 with minus but well perhaps there is something better in English to explain the variation -method. I am not sure whether it is the variation-of-parameters -method or something else, I cannot yet understand why this method is actually needed in my earlier question (I got the solution by integrating-factor -method -- or so I think, I may be wrong. Please, correct me if I am wrong.).

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• Hope my answers are helping you out!
– Pedro
Feb 20, 2012 at 15:17
• @Peter: yes, they do. I am slowly reading them and looking at the reference material every now-and-then, a bit slow but learning is. Thank you.
– hhh
Feb 20, 2012 at 18:20

Let's look at the second degree equation

$$y''+Py'+Qy=F(x) \tag{1}$$

where $P = P(x)$ and $Q = Q(x)$. Suppose we know the complementary solution to the ODE is

$$y = A u(x) + B v(x) \tag{2}$$

and now assume $A = A(x)$ and $B = B(x)$ are unknown functions to be determined, so that $y$ is a solution to the original ODE (Legendre's idea). We start our first differentiation

$$y' = A'u\left( x \right) + B'v\left( x \right) + Au'\left( x \right) + Bv'\left( x \right) \tag{3}$$

We assume our first equation:

$$A'u\left( x \right) + B'v\left( x \right)=0 \tag{4}$$

so from (3)

\begin{align*} y' &= Au'\left( x \right) + Bv'\left( x \right) \\ y'' &= Au''\left( x \right) + Bv''\left( x \right) + A'u'\left( x \right) + B'v'\left( x \right)\end{align*}\tag{5}

and replacing (5) to (1)

$$Au''\left( x \right) + Bv''\left( x \right) + A'u'\left( x \right) + B'v'\left( x \right) + P\left( {Au'\left( x \right) + Bv'\left( x \right)} \right) + Q\left( {Au\left( x \right) + Bv\left( x \right)} \right) = F(x).$$

Since $y$ is solution to the homogeneous equation, we have that

$$Au''\left( x \right) + Bv''\left( x \right) + P\left( {Au'\left( x \right) + Bv'\left( x \right)} \right) + Q\left( {Au\left( x \right) + Bv\left( x \right)} \right) = 0 \tag{7}$$

so giving the first equation to the system

\begin{align*} A'u'\left( x \right) + B'v'\left( x \right) &= F(x) \cr A'u\left( x \right) + B'v\left( x \right) &= 0 \end{align*}. \tag{8}

From algebra we have that this system is solved in terms of the determinants, which give

\eqalign{ & A' = \frac{{ - vF}}{{uv' - u'v}} \cr & B' = \frac{{uF}}{{uv' - u'v}} \cr}

Recalling the Wronskian Determinant of $u$ and $v$ is $W(u,v) = uv'-u'v$ we can write this as

\eqalign{ & A' = \frac{{ - vF}}{{W\left( {u,v} \right)}} \cr & B' = \frac{{uF}}{{W\left( {u,v} \right)}} \cr}

Which upon integration give

\eqalign{ & A = - \int {\frac{{vF}}{{W\left( {u,v} \right)}}} dx +C_1 \cr & B = \int {\frac{{uF}}{{W\left( {u,v} \right)}}} dx +C_2 \cr}

and make our solution be

$$y = v\left( x \right)\int {\frac{{uF}}{{W\left( {u,v} \right)}}} dx - u\left( x \right)\int {\frac{{vF}}{{W\left( {u,v} \right)}}} dx + C_1v\left( x \right) - {C_2}u\left( x \right)$$

I like to write this succintly as

$$y = v\int {\frac{{u \cdot F}}{W}} dx - u\int {\frac{{v \cdot F}}{W}} dx + C_1v - C_2u$$

Note that $C_1v - C_2u$ is the original solution to the homogeneous equation we had found.

• @hhh Yes it is. I usually refer to them has $y_c$, $y_p$, so that $$y_g = y_c+ y_p$$ i.e: "The general solution to an ODE is the homogeneous solution plus a particular solution to the non homogeneous equation.
– Pedro
Feb 23, 2012 at 20:44
• ...now working on (7) and (8) points, investigating...
– hhh
Feb 23, 2012 at 22:31
• Yes but that is the assumption? It relaxes the earlier assumption that $A(x)$ is a variable so it is a relaxation of the Legendre (Legendre's idea was to make a constant $A$ into $A(x)$ so I have understood from your writing), I think I kept the essence. The "assume" can be otherwise interpreted as unjustified premise -- I tried to clarify it, after thinking it. I find this as a weak link, could you clarify it -- why can you make such assumption?