# second order ODE via variation of parameters

I don't understand this step on this article. $$A'(x)u_1(x) + B'(x)u_2(x) = 0$$ why we desire A=A(x) and B=B(x) to be of this form? What is the basis that this form is valid?

-
So you ask "why". Well, just because it is useful. Have you read the entire article? Just go on, you will see what I mean and you will note how that condition is extremely useful. – Romeo Jul 25 '12 at 21:15
Yes i read the entire article for second order ...the whole problems is solved based on this. but why this condition holds? – Santosh Linkha Jul 25 '12 at 21:22
Well, the condition does not "hold". You simply require that A(x) and B(x) satisfy that condition. Seems incredible but it works and it simplifies next calculations; and, of course, it allows you to solve the algebraic system. Hope this helps – Romeo Jul 25 '12 at 21:26

This is closely tied to the method of osculating parameters. Suppose we wish to represent, with constant coefficients, some arbitrary function $u(x)$ with two linearly independent functions $u_1(x)$ and $u_2(x)$, $$u(x) = A u_1(x) + B u_2(x).$$ In general this can not be done. The best we can do is match the value of the function and its derivative at some point $x_0$, $$\begin{eqnarray*} u(x_0) &=& A u_1(x_0) + B u_2(x_0) \\ u'(x_0) &=& A u_1'(x_0) + B u_2'(x_0). \end{eqnarray*}$$ The conditions above determine the osculating parameters, the constants $A$ and $B$. $A$ and $B$ will be different depending on the point $x_0$. In general this fit will be poor at points far from $x_0$.

The method of variation of parameters involves finding the osculating parameters $A$ and $B$ at every point. That is, we let $A$ and $B$ be functions of $x$. The condition that they are the osculating parameters is that they satisfy $$\begin{eqnarray*} u_G(x) &=& A(x) u_1(x) + B(x) u_2(x) \\ u_G'(x) &=& A(x) u_1'(x) + B(x) u_2'(x), \end{eqnarray*}$$ just as above. For the second equation to hold it must be the case that $$A'(x)u_1(x) + B'(x)u_2(x) = 0.$$

-

Here’s the assumption. Simply to make the first derivative easier to deal with we are going to assume that whatever $u_1(t)$ and $u_2(t)$ are they will satisfy the following [...]