- #1

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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

- Thread starter asdf1
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- #1

- 734

- 0

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

- #2

saltydog

Science Advisor

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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.asdf1 said:

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

- #3

HallsofIvy

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?? Thereasdf1 said:

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

I THINK what you are talking about is "reduction of order". Suppose y

Then y'= u'y

I am, of course, using the "product rule". Notice that in the last term of both y' and y" I have only differentiated the "y

Now plug that into the equation:

(u"y

Combine the same derivatives of u:

u"y

Now, that u (as opposed to u' and u") is "gone" from equation

y"

u(y"

You now have u"y

- #4

saltydog

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I think he meant gone in equation 2 or at least that's how I interpreted it.HallsofIvy said:?? Therenever wasa "u" in equation 1- I certainly would say it was "gone"!

- #5

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sorry, i didn't write my question clearly...

i am talking about reduction of order and i meant gone in equation 2~

thank you! :)

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