# Derivation of Variation of Paramters

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First time poster, so please feel free to leave any comments of a general nature.

I'm hoping to get a further insight on the derivation of the variation of parameters method used in ordinary differential equations to solve linear second order equations. I understand were looking for a particular solution of the form
$$y_{p}=v_{1}(t) \cdot y_1(t)+v_2(t) \cdot y_2(t)$$
Taking the derivative with respect to t yields
$$\dfrac{d}{dt}y_p=(y_1 \cdot \dfrac{d}{dt}v_1 + y_2 \cdot \dfrac{d}{dt}v_2) + (v_1 \cdot \dfrac{d}{dt}y_1 + v_2 \cdot \dfrac{d}{dt}y_2)$$
The next step is to impose the requirement that
$$y_1 \cdot \dfrac{d}{dt}v_1 + y_2 \cdot \dfrac{d}{dt} v_2 = 0$$

It is at this point where I am not understanding (actually, this is the only step I don't understand in the derivation) . While I see that having the second derivatives of the functions [tex] v_1 [/itex] and [tex] v_2 [/itex] is going to be problematic, I do not understand why this requirement can be imposed without further explanation beyond "To simplify computation and to avoid second-order derivatives for the unknowns v1 and v2in the expression y''_p, we impose the requirement" (looked at a couple derivations online and in my old textbook, quote is from my textbook right before they impose the requirement I don't understand). Perhaps it is beyond the scope of the course, or perhaps I'm overlooking something blatantly simple? Any insight is appreciated.

2. ### matematikawan

333
Hi blinktx411, Welcome to the Differential Equations forum.

Remember that you are looking for a particular solution of the form yp=v1y1 + v2y2 .
You have two unknown functions v1 and v2 to be determined. But how many known equations do you have to solve this? Only one, from the given DE.

So you are free to do with what you like for the second equation as long as you are able to solve for v1 and v2 . (remember that you are just looking for a particular solution).

It took the genius of Lagrange to identified that second equation. And the method also works nicely with higher order.