Question

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

theres a long way and an even longer way

Answer 2

a wronskian appraoch is prolly the simplest and shortest longer way

Answer 3

whats our homogenoues solution using the characteristic eq?

Answer 4

can you explain what you mean by using the wronskian? just to find the independent solutions?

Answer 5

i can, but its really the result of using variation of parameters, the longer method

Answer 6

first find the homog part assuming x^r is a solution

Answer 7

y''+[(1/x)*(y')]-[(1/x^2)*y]=ln(x)

rewrite it as;

x^2 y''+x y'- y = x^2 ln(x)

Answer 8

assuming x^m is a solution to the couchy

Answer 9

ok im following

Answer 10

y = x^m
y' = x x^(m-1)
y'' = x(x-1) x^(m-2)

sub those in to find the homog solution, and our y1 y2 generals that we will need in the wronskian

Answer 11

then we set up the Wronskian, or rather i set it up, like this and take a pseudo cross
\[\begin{vmatrix}W_1&W&W_2\\y_1&0&y_2\\y'_1&ln(x)&y'_2\end{vmatrix}\]

Answer 12

\[W_1=-y_2ln(x)\]\[W=y_1y'_2-y_2y'_1\]\[W_2=y_1ln(x)\]

our particular solution then turns out to be
\[yp_1 = y_1\int\frac{W_1}{W}\]\[yp_2 = y_2\int\frac{W_2}{W}\]

Answer 13

Can you explain why you set it up that way? isn't a Wronskian supposed to look like this? \[\left[\begin{matrix}y1 & y2 & y3 \\ y1' & y2' & y3' \\ y1(n-1) & ... & y3(n-1) \end{matrix}\right]\]

Answer 14

my set up is only a means to organize the information

Answer 15

and no, your example is not the Wronskian method

Answer 16

similar yes, but not organized correctly

Answer 17

I didn't learn that way yet :\

Answer 18

the Wronskian is the cramer rule result of doing it the longer way

Answer 19

I see, I will have to read up on those two methods since we haven't covered them yet. Do you have any good resources or suggestions for reading?

Answer 20

i havent seen any one set it up like i do, simply because i develop my own style. but this usually is coherent enough to follow

http://tutorial.math.lamar.edu/Classes/DE/Wronskian.aspx

Answer 21

Ok thank you! I will try to follow that approach and see if I can translate it to the problem.. But you say this is possible with variation of parameters as well?

Answer 22

yes, since this is just a restructring of the variation of parameters

Answer 23

http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/ode/second/so_var/so_var.html

notice their use of the W

Answer 24

one more question - why did you multiply by x^2 in the beginning?

Answer 25

in order to solve the homogenous part; we dont need the fractions

Answer 26

The Wronskian is related!!! I see, we didnt learn that yet.. hum

Answer 27

we can go thru the longer version just as well

Answer 28

Wait I'm still trying to understand everything you did, since I've seen a similar form used briefly by my teacher but not well explained - are you assuming a power law when you write:

y = x^m; y' = (x)(x^m-1); y'' = (x-1)(x^m-2)

Answer 29

yes

Answer 30

And then you plug into the original differential to use the W approach?

Answer 31

the functions of x as coeefs tells me this is a couchy form of a diffyQ

Answer 32

\[x^2(m(m-1))x^{m-2}\]\[+x(m-1)x^m\]\[-x^m\]

\[x^m(m^2-m+m-1-1)=0;\ m^2=0\]

we have a double root it looks like so:\[y=c_1x^0+c_2x^0ln(x)+yp\]

Answer 33

Ok - just read the wiki and I follow your approach - but how to solve for yp then?

Answer 34

Using this approach?

Answer 35

assume c1 and c2 are functions of x; A and B

\[yp=A+Bln(x)\]
\[yp'=A'+B'ln(x)+\frac{B}{x};\ A'+B'ln(x)=0\]\[yp'=\frac{B}{x}\]
\[yp''=\frac{B'}{x}+\frac{xB'-B}{x^2} \]

with any luck, when we use these in the original, things get simplified :)

Answer 36

\[x^2(\frac{B'}{x}+\frac{xB'-B}{x^2})+x(\frac{B}{x})-(A+Bln(x))]=x^2 ln(x)\]
\[B'x+xB'-B+B-A-Bln(x) =x^2 ln(x)\]
\[2B'x-A-Bln(x) =x^2 ln(x)\]

seems im rusty on the longest of the longer version lol

Answer 37

\[y _{h}=[x ^{2}(m-1)(x^{m-2})]+[x(m*x^{m-1})]-x^{m}=0\]

Answer 38

thats what I get for the homogenous solution - which is different than your solution -_-

Answer 39

you didnt derive x^m correctly for each spot

Answer 40

x^m; m*x^m-1; and (m-1)(x^m-2)

Answer 41

for y y' and y''

Answer 42

\[y=x^m\]\[y'=m\ x^{(m-1)}\]\[y''=m(m-1)\ x^{(m-2)}\]

Answer 43

ohhh i forgot an m

Answer 44

we are just using the power rule

Answer 45

Ok so then you factor x out and make an expression for m - then solve for the roots of m to get your homogenous solution right?

Answer 46

yes

Answer 47

how do you group the terms that have x to some m power? - i dont think i've factored an expression like this before

Answer 48

I get x^m+1 on the outside - but idk if im diong it right

Answer 49

\[x^2 * x^{m-2}=x^{m-2+2}\to\ x^m\]

Answer 50

all the xpart go away and we are left with an x^m to factor out that never goes zero so we can ignore it

Answer 51

m(m-1) + m - 1 = 0

Answer 52

m^2 - 1 = 0 when m=1 or -1

Answer 53

looks like a mismathed at the start of this post also :)

Answer 54

yes i thought it was just a typo but i guess it carried to the solution

Answer 55

this is what i have so far worked out

Answer 56

\[y_{h}=x^{m}(m(m-1))+x^{m-1}(m)-x^{m}\]

Answer 57

(btw do you know if this board uses latex for equations?)

Answer 58

it codes latex

Answer 59

ohh thats awesome - bc ive been meaning to learn, now I will have to use this all the time (my first time on this board)

Answer 60

your middle should be x^m as well; not x^(m-1)

Answer 61

\[x(y')=x(m*x^{m-1}) since y=x^{m} and y'=m*x^{m-1}]

Answer 62

x*x^(m-1) = x^m

Answer 63

er\[(x(y')=x(m*x^{m-1}) since y=x^{m} and y'=m*x^{m-1}\]

Answer 64

\text{} writes in latex

Answer 65

ah ok noted

Answer 66

\[this and that\text{ : this and that}\]

Answer 67

\[x(y′)=x(m∗xm−1)\text{ since, }y=xm\text{ and, }y′=m∗xm−1\]

Answer 68

forgot to put in exp when copy paste

Answer 69

lol

Answer 70

I got x^m now though

Answer 71

x*x^{m-1} = x^{m} right?

Answer 72

yes, simplifying to m x^m

Answer 73

Yup - ok so the general solution to this then is y = Cx^m

Answer 74

yes, but:\[y=c_1x^{-1}+c_2x^1+yp\]

Answer 75

so now i can use variation of parameters or solve for the coefficients

Answer 76

yes

yp = Ax^-1 + Bx
yp' = A'x^-1 -Ax^-2 +B'x + Bx' ; the A' B' stuff sums to zero

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

Answer 77

yup yup i see

Answer 78

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

Answer 79

now we setup a system of eqs;

we assumed; A'/x + B' x = 0
and now; A'/x + B' x = x^2 ln(x)

Answer 80

solve for A' B' and then integrate them up tp A and B

Answer 81

how does A'x^-1 go to zero in :

yp' = A'x^-1 -Ax^-2 +B'x + Bx' ; the A' B' stuff sums to zero

Answer 82

A'x^-1 + B'x = 0 ... yes but can you add them?

Answer 83

since c1 x^-1 + c2x = 0; its really just the result of the yh part since c1=A and c2 = B

Answer 84

so yh'=0 then?

Answer 85

yes, thats the definition of homogenous equations; this = 0

Answer 86

and since c1 is arbitrary and c2 as well; A' and B' makes no difference; it just goes to zero

Answer 87

ah subtle - but straightforward, ok i believe you

Answer 88

ok still working through it, trying to solve for coeffs

Answer 89

as a result, we can cancel them from the yp part to determine the particular results

Answer 90

Ok, so I got this: x^2 ln(x) = -A' + B'x^2 --> but you said that:

we assumed; A'/x + B' x = 0 and now; A'/x + B' x = x^2 ln(x)

Answer 91

one is a result of the yh part ; which equals 0

the other is a result of the yp part; which equals x^2 ln(x)

Answer 92

So i use these two equations to solve for the coefficients? What about the x^2*ln(x) = -A' + B'*x^2

Answer 93

what was the point of getting tha result?

Answer 94

in order to put it all together since the general solution is the sum of the yh and the yp parts

Answer 95

but we have to determine A and B for the general solution, not A' and B'; so we have to solve this system for A' and B' in order to integrate A' to A, and B' to B

Answer 96

Yes i understand that part but now I am confused with which equations to use to solve for A' and B', it is not

A'x+B'x = 0
-A'+B'x^2 = ln(x)*x^2

Answer 97

i was just checking the wolf to make sure we was on the right track

Answer 98

Ok - i hope it is right, bc ti makes sense to me ha