Question

HELP!!! Reduction of Order Question.

Answer 1

\[y1(x) = 1/(1-x) \]
\[x(x-1)y'' + (1-3x)y' -y = 0\]
Use reduction of order to find another inderpendent linear solution

Answer 2

\[y_2 = y_1 (u) = u/(1-x)\]
\[y'_2 = \frac{ (1-x)u'+u }{ (1-x)^2 }\]
\[\frac{ 2[(1-x)u'+u] }{ (1-x)^3 }-\frac{ u'' }{ (1-x) }\]

Answer 3

and then I subtitute all these 3 into the ODE, where basically my brain has stopped working for this part as the equation doesn't make sense to me
\[x(x-1)[\frac{ 2[(1-x)u'+u] }{ (1-x)^3 }- \frac{ u'' }{ (1-x) }] +(1-3x)\frac{ (1-x)u'+u }{ (1-x)^2 }-\frac{ 1 }{ (1-x) }\]
after simplify it,
\[-u'' + [\frac{ 2x }{ (1-x) }+\frac{ (1-3x) }{ (1-x) }]u'+u[\frac{ 2x }{ (1-x)^2 }+\frac{ (1-3x) }{ (1-x)^2 }-\frac{ 1 }{ (1-x) }]\]
After this... I have no idea how to continue this anymore.

Answer 4

@phi can you help?

Answer 5

@TheSmartOne can you help?

Answer 6

@ParthKohli can you help?

Answer 7

@zepdrix Can you help?

Answer 8

shouldn't you start with
\[ y_2= y_1 \int u(x) \ dx \\ y_2 = \frac{1}{1-x} \int u(x) \ dx\]
?

Answer 9

also, your y1 is not a solution to the diff. eq. so I suspect there is a typo in the question
if the original equation were

x(1-x)y'' + (1-3x)y' -y = 0

(note the (1-x) instead of (x-1) )
then y1= 1/(1-x) is a solution

Answer 10

y1=1/(1-x) is not the solution to that differential equation .

I also think that there is a typo in the equation . it should be (1-x) there

1)take y2=u.y1 , where u is some function of x , then substitute directly in the differential equation and use product rule . there will be lot of cancellations as y1 is a solution

2) After simplifying you would probably get a 1st order equation , that you can solve !

Please try and reply . meanwhile i will try solving it :)

Answer 11

with any luck you should get
y2= ln(x)/(1-x)

Answer 12

unless I misunderstand the question here, otherwise y_1 = 1/(1-x)

Answer 13

ok, that looks better. The equation you posted at the very top has a typo.

they first want you to verify that 1/(1-x) is a solution. I assume you have done that ?

to find a 2nd solution, let \[ y_2 = y_1 \int u(x) \ dx \]
the derivatives are
\[ y_2 '= y_1 u + y_1' \int u(x) \ dx \\ y_2''= y_1 u' +2 y_1' u + y_1'' \int u(x) \ dx \]

Answer 14

if we consider the general equation
\[ f_2(x) y'' + f_1(x) y' + f_0(x) y = 0 \]
and plug in \(y_2\) and its derivatives
we will get
\[ \left(f_2(x) y_1'' +f_1(x) y_1' +f_0(x) y_1\right) \int u(x) \ dx + \\ f_2(x)y_1 u' +[2 f_2(x) y_1'+f_1(x)y_1]u =0\]
the first line (because f1 is a solution) is zero, and disappears...
thus, after substituting you should get just the 2nd line, which you solve for u

Answer 15

in your problem, f2 is x(1-x) and f1 is (1-3x)

Answer 16

**the first line (because y1 is a solution) is zero, disappears

Answer 17

thank you so much for the help. I just want to make sure for finding the u.
I basically subs. in f_2(x) and f_1(x) including y_1 and y_1' as I assume y_1 = 1/(1-x) consider I have already verified 1/(1-x) is a solution. So, is it how I do it?

Answer 18

yes, and you get a (simple) differential equation for u that you solve
once you have u, you can find the second solution y2 by integrating u
and then multiplying by that result by y1

Answer 19

****how do I verify the that 1/(1-x) is a solution? ***
you find its first and 2nd derivatives and "plug them in" to the equation and verify the equation is true.

Answer 20

Thank you so much. It's all done now. :)