Why is 'u' gone in equation 1?

[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:
u``y1 + u`(2y`1+py1) + u(y``1+py`1+qy1)=0 ...equation 2
and y1 is a solution of equation 1
then why is "u" gone in equation 1?
:P

 

[saltydog]

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:
u``y1 + u`(2y`1+py1) + u(y``1+py`1+qy1)=0 ...equation 2
and y1 is a solution of equation 1
then why is "u" gone in equation 1?
:P

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.

 

[HallsofIvy]

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:
u``y1 + u`(2y`1+py1) + u(y``1+py`1+qy1)=0 ...equation 2
and y1 is a solution of equation 1
then why is "u" gone in equation 1?
:P

?? 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.

 

[saltydog]

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

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

 

[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! :)