Couldn't understand the proof for Method of variation of parameters

[iVenky]

Here's the proof that I read for method of variation of parameters-

https://www.physicsforums.com/attachment.php?attachmentid=52267&stc=1&d=1351081780

What I couldn't understand is that how could one simply assume that

u'₁y₁+u₂'y₂=0 and
u'₁y'₁+u₂'y'₂=g(x)

I just don't understand from where you get those above two results.
I would be really happy if could clear my doubt.

Thanks a lot for physics forum and its members. You have helped me a lot :)

 

[Vargo]

In a nutshell, you assume that the first equation is true. That assumption is part of the technique. Once you assume that the first equation is true, you plug u1y1+u2y2 into your original differential equation and you end up with the second equation.

Think of it this way, you have two unknowns (u1, and u2). But your original differential equation is only one equation. So we just set u1'y1+u2'y2=0 so that we have two equations for two unknowns.

With that said, a second order ODE can be converted into a first order system of ODE for two unknown functions. If you do variation of parameters on THAT system, then you get the same two equations without having to make up one of them out of thin air. So in that sense, the first equation is not really just an arbitrary choice that makes it easy to solve for u1 and u2. It can be derived from something that comes from a later topic in your course.

 

[iVenky]

In a nutshell, you assume that the first equation is true. That assumption is part of the technique. Once you assume that the first equation is true, you plug u1y1+u2y2 into your original differential equation and you end up with the second equation.

I did what you said. I assumed the first one and tried to plug that into the first equation which is

y''+p(x)y'+q(x)y=g(x)

but I couldn't get the second equation.

Here's my derivation-

y=u₁y₁+u₂y₂
y'=u₁y'₁+u₂y'₂

as we have assumed that u'₁y₁+u'₂y₂=0

y''=u'₁y'₁+u'₂y'₂+u₁y''₁+u₂y''₂

and if we plug in these expressions for y,y',y'' in the equation


y''+p(x)y'+q(x)y=g(x)

I don't get the second equation. I mean there are many terms apart from
u'₁y'₁+u'₂y'₂

and how did you neglect them?

Thanks a lot :)

 

[Vargo]

Those terms cancel each other out. You have to use the fact that y1 and y2 each satisfy the homogeneous equation.
y1''+py1'+qy1=0, and the same for y2.

 

[iVenky]

Those terms cancel each other out. You have to use the fact that y1 and y2 each satisfy the homogeneous equation.
y1''+py1'+qy1=0, and the same for y2.

Got it.

Thanks :)