OpenStudy (anonymous):

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.

3 years ago
OpenStudy (anonymous):

\[xy"+y'=o\] solution \[y_1 (x)=1\]

3 years ago
OpenStudy (accessdenied):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (accessdenied):

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..

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (accessdenied):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (accessdenied):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (accessdenied):

yup, i can do that. :)

3 years ago
OpenStudy (anonymous):

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?

3 years ago
OpenStudy (anonymous):

Sorry dv/dx

3 years ago
OpenStudy (accessdenied):

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?

3 years ago
OpenStudy (accessdenied):

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

3 years ago
OpenStudy (accessdenied):

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 \( (x v)' = 0\)

3 years ago
OpenStudy (accessdenied):

"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.

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (accessdenied):

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

3 years ago
OpenStudy (anonymous):

so y'=c1/x?

3 years ago
OpenStudy (accessdenied):

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.

3 years ago
OpenStudy (anonymous):

Sorry I'm not following you...

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (anonymous):

sorry xy"+c1/x=0

3 years ago
OpenStudy (accessdenied):

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.

3 years ago
OpenStudy (accessdenied):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
OpenStudy (accessdenied):

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?

3 years ago
OpenStudy (accessdenied):

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

3 years ago
OpenStudy (anonymous):

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

3 years ago
Similar Questions: