Question

a differential equation and one solution are given use d'alembert's reduction of order method to find a second linearly independent solution what is the general solution of the differential equation.

Answer 1

\[xy"+y'=o\]

solution

\[y_1 (x)=1\]

Answer 2

so do you know what is meant by d'alembert's reduction of order?

Answer 3

not really I've been looking at the book and I can't figre it out

Answer 4

essentially, the concept is to look at the derivatives in your equation. note that there are no plain ol' y terms. they're all derivatives..

Answer 5

I have d'alembert's equation if that would help

Answer 6

so we can make a substitution like this: v = y'
it will "reduce" the order of the DE because : v = y' v' = y''

Answer 7

so I get xv'+v=0 and just integrating factor method?

Answer 8

yup. once you solve for v, you go back to this equation: v = y' <-- and integrate this to solve for y again.

Answer 9

Ok let me try that out would you mind checking the work?

Answer 10

yup, i can do that. :)

Answer 11

ok I hit a jam earlier than I thought I have never done a problem with a coeffient attached to my y' term, but simple algebra says I can say y'=-v/x, and normally the y'=dy/dx in this case would it be dy/dv?

Answer 12

Sorry dv/dx

Answer 13

hm.. i can't quite determine where you are heading with that.
\( x v' + v = 0 \)
we could divide off the x from both sides to get rid of that coefficient
\( v' + \dfrac{1}{x} v = 0 \)
which looks like more familiar territory?

Answer 14

the y should not come into play until we solve this DE for the solution of v, if that is clear.

Answer 15

Or a bit of a short-cut, if you notice the format of the initial equation and x. x'=1
\( x v' + (x)' v = 0 \) <-- also product rule
\( (x v)' = 0\)

Answer 16

"but simple algebra says I can say y'=-v/x, and normally the y'=dy/dx in this case would it be dy/dv?"
Oh.
xv' + v = 0
v' + v/x = 0
v' = -v/x
I got mixed up seeing the y in there. This also separates. v' = dv/dx.

Answer 17

ok I worked it out and solved for v and get v=c1/x

Answer 18

that looks good to me. :)
so for y? v = y'

Answer 19

so y'=c1/x?

Answer 20

correct, we just have to solve then for y.
it would be easy to just integrate both sides, or if you felt more comfortable, separate variables works again.

Answer 21

Sorry I'm not following you...

Answer 22

wouldn't our new expression be xy"=c1/x=0

Answer 23

sorry xy"+c1/x=0

Answer 24

we know that y' = c1/ x is true so far. so it should work in our equation naturally. but then we will just get an invariably true statement by putting it back into the equation (note that your y" = d/dx (c1/x) also is put in)
we want to find a second linearly independent solution using the general form, so solving this: y' = c1/x will give us the solution y in terms of x.

Answer 25

if you're still uncertain, i invite you to test it out for yourself on some scrap paper:
\( y' = \dfrac{c_1}{x} \) and \( y'' = \dfrac{d}{dx} \left( \dfrac{c_1}{x} \right) = - \dfrac{c_1}{x^2}\)
by substituting this, you get in the original equation 0 = 0

Answer 26

Ok sorry to be a pest but I'm confused on how to get from where we are to the answer the book has which is y=c1+c2Ln, the book is wrong pretty often though

Answer 27

nope, you're fine.
think of this as almost a totally separate differential equation now:
\( y' = \dfrac{c_1}{x} \)
how would you solve it for y?

Answer 28

you could again rewrite y' = dy/dx
\( \dfrac{dy}{dx} = \dfrac{c_1}{x} \)

if that helps.

Answer 29

Thank you, let me see if I can figure out the next one