Question

Solve the initial-value problem (x-1)y'+3y=1/(x-1)^3+sin(x)/(x-1)^2, y(0)=1.

Answer 1

I don't know why the answer is y=(x-1)^(-3)*[ln(1-x)-cos(x)] when I got the answer y=(x-1)^(-3)*[ln(x-1)-cos(x)].

Answer 2

Please help.

Answer 3

y=(x-1)^(-3)*[ln(1-x)-cos(x)]
y=(x-1)^(-3)*[ln(x-1)-cos(x)]
when x=0, ln(-1) is undefined

its prolly a simply error if an error at all

Answer 4

So what's the correct answer? I got ln(x-1) instead of ln(1-x) because it's ln abs(x-1) so when x=0, ln(-1) becomes ln(1)=0, isn't it? Because of the absolute value.

Answer 5

then id say its just a formating issue and that ln|x-1| is mathematically fine

id have to review your process tho

Answer 6

Let me show my work. Please wait. It won't be long.

Answer 7

y'+3/(x-1)y=1/(x-1)^4+sin(x)/(x-1)^3

u=e^(3*ln abs(x-1))=(x-1)^3

(x-1)^3*y'+3(x-1)^2*y=1/(x-1)+sin(x)

(x-1)^3*y=ln abs(x-1)-cos(x)+C

y=(ln abs(x-1)-cos(x)+C)/(x-1)^3

y(0)=(-1+C)/-1=1

C=0

y=(x-1)^-3*[ln abs(x-1)-cos(x)]

Answer 8

does this look correct?

u = e^(3*ln abs(x-1)) = |x-1|^3

Answer 9

Yes.

Answer 10

So what's the right answer?

Answer 11

both formats are correct, as far as i can see.

Answer 12

So my answer is correct?

Answer 13

you can always double check it by plugging it into the setup :) but yeah, one has a restricted domain, whereas the other does not; but they both graph out the same.

Answer 14

Thank you!

Answer 15

http://www.wolframalpha.com/input/?i=ln%28%7Cx-1%7C%29-ln%281-x%29%3D0

as far as the restrictions go, they are the same function for x<1, which x=0 is definantly <1

Answer 16

Thank you so much for the extra info! Now I understood this problem!

Answer 17

good luck :)

Answer 18

:)