Question

seldamat

Solve the differential: y''+[(1/x)*(y')]-[(1/x^2)*y]=ln(x) What would be the best way to approach a problem like this? Variation of parameters?

Answer 1

you stated from the last one:"I was thinking to solve, would be to write out x^2*yp''+x*yp'+yp = x^2*lnx as we solved for them - then equate the coefficients to the right hand side"

much better now

Answer 2

this would have been sufficient for constant coeffs i beleive

Answer 3

yes, i am trying that now,

Answer 4

nm .. lol sooo should I just plug in then what?

Answer 5

but since our "coeffs" are functions, that wont be sufficient for our needs

Answer 6

recall that we solced the homogenous part: and im going to forgo the latex since thats alot of the slowdown

yh = c1 x^-1 + c2 x = 0

Answer 7

since our coeffs are functions of x, and not constants; we let c1 and c2 equal arbitrary functions in x and equate this to the yp part

Answer 8

yup! i follow

Answer 9

yp = a(x) x^-1 + b(x) x

yp' = -a(x) x^-2 + a'(x) x^-1
+b'(x) x + b(x)x'
-----------------------------
-a(x) x^-2 + 0 + b(x)

good so far?

Answer 10

yes i have that

Answer 11

yp'' = 2a(x) x^-3 - a'(x) x^-2 + b'(x)

plug these values into the eq to simplify it such that it equates to x^2 ln(x)

Answer 12

yup that is what im working on now

Answer 13

yup so i get teh same answer= -A' + B'x^2 = x^2*ln(x)

Answer 14

x^2 yp'' = 2a(x) x^-1 - a'(x) + b'(x) x^2
+ x yp' = -a(x) x^-1 + b(x) x
- yp = -a(x) x^-1 -b(x) x
---------------------------------
x^2 ln(x) = - a'(x) + b'(x) x^2

Answer 15

good, now we have 2 equations in A' and B' to solve for a system of eqs

Answer 16

yh and the equation above?

Answer 17

yes

Answer 18

so yh = A'x^-1 + B'x = 0

Answer 19

yes, and with any luck our longest version of this method is accurate lol for our yp in A' and B'

Answer 20

sooo i got B' = ln(x) / 2

A' = (-ln(x)/2)^2

Answer 21

er minus on the outside

Answer 22

A' = -(ln(x)/2)^2 B' = ln(x)/2

Answer 23

if that looks right im gonna try and solve for y

Answer 24

you have to find A and B so integrate thos

Answer 25

yup, just realized that as i started to plug in

Answer 26

i can show u how to do it by variation of parameters if u want,, i think its easier in this case

Answer 27

this is variation of paramenters; the long version

Answer 28

he hasnt gotten to the wronskian method yet

Answer 29

oh lol i saw the letters i assumed its undetermined coeffecients,, never mind:)

Answer 30

the wronskian sounds easier (still working, solving for A now - had to brush up on integration by parts lol)

Answer 31

i had to go to paper to keep track of things :)

i get:

a' = -x^2 ln(x)
b' = ln(x)/(1-x)

Answer 32

|a' = |-x^2 ln(x) = -ln(x) x^3/3 - | -x^3/3x
= -ln(x) x^3/3 +| x^2/3

\(\large = \frac{x^3(1-3ln(x))}{9}\)

Answer 33

the wolf says we end up with:

y = c1 x^-1 + c2 x - ln(x)

Answer 34

i got something different

Answer 35

still simplifying though

Answer 36

yeah, my mathing is off somplace i think

Answer 37

the wronskian is way way better for this mess :/

Answer 38

y = c1x^-1 + c2x + ln^2(x) - 2ln(x) + (x^2*ln(x)-x) / 2

Answer 39

where A' = ln^2(x)/4 and B' = ln(x) / 2

Answer 40

B = 1/2*(ln(x) - x)

A = [ln^2(x)]*x - integrand(x*2ln(x)*1/x dx) (integration by parts)

solve and you get

A = ln^2(x)*x - 2*x(ln(x) -1 )

Answer 41

dbl chk your system of eqs for a' and b'

Answer 42

yea i'm pretty sure that they are right

Answer 43

A' and B' that is

Answer 44

\[\begin{align}x^{-1}A'+xB' &= 0\\-x^{-1}A'+1B' &= ln(x)\end{align}\]

Answer 45

(1+x)B' = ln(x)


B' = ln(x)/(1+x)

Answer 46

x^-1A' = -xB'

A' = -x^2 B'

Answer 47

how did you get -A'/x + B = ln(x)

Answer 48

that was the end results of the plugging and playing with the yp and its derivatives

Answer 49

[-A'+2A/x + B'x^2] + [-A/x + B/x] - [A/x + Bx] = x^2*ln(x)

Answer 50

-A' x + B'x^2 = x^2 ln(x) ; divide off the x^2

-A'/x + B' = ln(x)

Answer 51

then -A' + 2A/x - 2A/x + B'x^2 + Bx - Bx = x^2 * ln(x)

Answer 52

-A'x at the start, see your typo

Answer 53

the first expression i sent is after you distribute the x^2 - once you do that, you lose the x on A'

Answer 54

x/x^2 = 1/x

Answer 55

-A' x/x^2 = -A/x

Answer 56

x^2(yp'') = x^2 ( 2Ax^-3 - x^-2A' + B')

Answer 57

yeah, i just seen that lol

Answer 58

prolly why my results were so ugly ...

Answer 59

haha its ok, my result is ugly too

Answer 60

a'/x + b' x = 0 <-- * -/x
-a'/x^2 + b' = ln(x)



-a'/x^2 - b' x = 0
-a'/x^2 + b' = ln(x)
--------------------
-2a'/x^2 = ln(x)

a' = -ln(x)/2

-ln(x)/2x^2 - b' x = 0

-ln(x)/2x - b' = 0

-ln(x)/2x^3 = b'

now whatd i miss on that :)

Answer 61

other than the x^2 .....

Answer 62

a' = -ln(x)/2 * x^2 right?

Answer 63

a' = -x^2 ln(x)/2


-x^2 ln(x)/2x + b' x = 0 : /x

-x^2 ln(x)/2x^2 + b' = 0

-ln(x)/2 + b' = 0

b' = ln(x)/2 .....maybe?

Answer 64

i have a theory; since i had to make room for the wronskian method; i had to forget alot of algebra in the process :)

Answer 65

i suck at alegbra... lol so I should have tons of room for the wronskian i hope (according to your theory :P )

Answer 66

i need to eat some lunch.. im out of brain juice, then ill com eback to check work and see if there are any obvious errors

Answer 67

i looked at the wolfram solution - it has imaginary numbers in it ...

Answer 68

ok brb before i starve myself

Answer 69

|b'= | ln(x)/2 = x ln(x)/2 - | x/2x
= x ln(x)/2 - x/2

|a' =|-x^2 ln(x)/2 = -x^3 ln(x)/6 + | x^3/6x
= -x^3 ln(x)/6 + x^3/18

a/x = -x^2 ln(x)/6 + x^2/18
+bx = x^2 ln(x)/2 - x^2/2
-----------------------------
yp = x^2/2 (1-3ln(x)+9ln(x)-9)


yp = x^2/2 (6ln(x)-8)
= x^2 (3ln(x)-4)

but typing and mathing fractions and such aint my forte so youd better double chk it :)

Answer 70

I checked it btw - and it worked like a charm. Thank you so much! I've learned a lot you know.