# Seems a little silly

1. Aug 14, 2005

### asdf1

This question seems a little silly, because it looks so simple
but
There's one thing I don't understand:
If y +p(x)y + q(x)y=0 ...equation 1.
and you have this equation:
uy1 + u(2y1+py1) + u(y1+py1+qy1)=0 ....equation 2
and y1 is a solution of equation 1
then why is "u" gone in equation 1?
:P

2. Aug 14, 2005

### saltydog

Because y1 satisfies the first equation and that equation is the coefficient of u in the second equation and since the first equation is set to zero when y1 is plugged into it, then the coefficient of u in the second one is zero.

3. Aug 14, 2005

### HallsofIvy

Staff Emeritus
?? There never was a "u" in equation 1- I certainly would say it was "gone"!

I THINK what you are talking about is "reduction of order". Suppose y1 is a solution of equation 1 and let y= u(x)y1 (x).

Then y'= u'y1 + uy'1 , y"= u"y1 + 2u'y'1 + uy"1 .

I am, of course, using the "product rule". Notice that in the last term of both y' and y" I have only differentiated the "y1" part- its as if u were a constant.

Now plug that into the equation:
(u"y1 + 2u'y'1 + uy"1)+ p(x)(u'y1 + uy'1)+ q(x)(uy)= 0.

Combine the same derivatives of u:
u"y1+ u'(2y'1+ p(x)y1)+ u(y"1+ p(x)y'1+ q(x)y1)= 0

Now, that u (as opposed to u' and u") is "gone" from equation 2 (not equation 1- that must have been a typo) because y1 satisfies the original equation:
y"1+ p(x)y'1+ q(x)y1= 0 so
u(y"1+ p(x)y'1+ q(x)y1)= u(0)= 0.

You now have u"y1+ u'(2y'1+ p(x)y1)= 0. If you let v= u', that becomes v'y1+ v(2y'1+ p(x)y1)= 0, a simple, separable, first order equation. Solve for v(x), integrate to find u(x) and form u(x)y1 to find the second, linearly independent solution.

4. Aug 14, 2005

### saltydog

I think he meant gone in equation 2 or at least that's how I interpreted it.

5. Aug 15, 2005

### asdf1

you're both right~
sorry, i didn't write my question clearly...
i am talking about reduction of order and i meant gone in equation 2~
thank you! :)