Question

Solve 2ty'+y=1

Answer 1

Here4 is what I got so far: \[2ty'=1-y\\y'=\frac{1-y}{2t}\\\frac{dy}{1-y}=\frac{dt}{2t}\]

Can I do that? There is no t doesn't =0 stipulation

Answer 2

where did you get 1-y

Answer 3

sorry , mistyped original fixed now

Answer 4

I mean, I can do it from there, I'm just wondering if the dividing by 2t is legal if t can =0

Answer 5

it's just separation of variables if I can divide by t. So for integrating factor it would beeee....h/o (I'm not so great with them so I need a minute)

Answer 6

would it not just be

Answer 7

yes you can divide by 2t, thats fine

Answer 8

divide by 2

Answer 9

oh, then I'm good doing separation of variables?

Answer 10

-ln|1-y| = 1/2 ln |t| + c

Answer 11

because y' is in terms of t so you have
tdy/dt= (1-y)/2

Answer 12

oh yeah i get it my bad

Answer 13

yea, I know, I just wanted to make sure I could divide by t like I need to

Answer 14

can you show me the integrating factor way @perl ? Is the integrating factor \(e^{-t^2}\)?

Answer 15

sure
$$\Large\rm {
2ty' + y = 1
\\y' + \frac{1}{2t}y = \frac{1}{2t}
\\ integrating ~ factor ~ e^{ \int \frac{1}{2t}}
}
$$

Answer 16

wait, I thought the integrating factor was on p(t)?

Answer 17

yes p(t) = 1/(2t)

Answer 18

but isn't p(t) the coefficient on the y' term? Then q(t) the coefficient on the y term?

Answer 19

first step is to divide through by the coefficeint of y '

Answer 20

oh, I thought that was only for second order

Answer 21

ohhhkkkk

Answer 22

integrating factor is used on :
$$
\Large \frac{dy}{dt} + p(t) y = g(t)
$$

Answer 23

...Ohk, now I get it.

Answer 24

Thank you!!

Answer 25

You can stop there, btw my final answer is \(y=1-c_1t^{-1/2}\) do you agree?

Answer 26

ah absolute value on that t

Answer 27

yes thats correct, but you don't need to put a negative in front of c1

Answer 28

ok thanks!!!

Answer 29

not that its wrong, but you could have a plus as well :)
since its a constant it will absorb sign

Answer 30

yea, I did that with the e^-c I just didn't mind the - since y(1)=0 and I'd need a negative anyways

Answer 31

thanks again Perl!!!

Answer 32

your welcome