Question

differential equation SOS
y'-y=2x-3

Answer 1

this is a linear differentiation equation that can be solved with a integrating factor

Answer 2

I don't get to the final answer

Answer 3

yeah, it's just the beginning

Answer 4

\[v=e^{\int\limits (-1) dx}\]
this is the integrating factor
-1 is up there becaue
your diff is in the form
y'+py=q
where we choose the integrating factor to be
\[v=e^{\int\limits p dx}\]

Answer 5

so can you integrate the -1 up there w.r.t. x?

Answer 6

we didn't learn that....

Answer 7

oh then what are you guys doing

Answer 8

like are you doing that like guessing type thing
where you try to determine some coefficients bases on what you think the solution will look like
or whatever they call it

Answer 9

Singular solutions

Answer 10

like putting dy/dx instead of y'

Answer 11

I don't think singular solutions is a method to solve the differential equation

Answer 12

maybe first order?

Answer 13

I don't get it.
Are we not suppose to solve the differential equation?

Answer 14

idk the title tbh, our teacher doesn't speak much the language, he does solve though

Answer 15

is there may be an easier way?

Answer 16

so if he said to solve
then you must know a method to solve this first order linear differential equation

Answer 17

It was our first lesson after all

Answer 18

and I need to know the method

Answer 19

well thanks though

Answer 20

I know two methods.
One is called using a integrating factor and the other is called something about undetermined coefficients. So neither of those ring a bell for you?

Answer 21

the first one yes!

Answer 22

out TA told us to switch y' with dy/dx

Answer 23

oh i thought you said you didn't know that method earlier

Answer 24

I was doing the whole integrating factor thing
and you know you didn't know that route remember?

Answer 25

I'll go though on it again @freckles by myself thanks though

Answer 26

I'll try again by myself first I think

Answer 27

\[y'+py=q \\ \text{ Integrating factor is } v=e^{\int\limits p dx} \\ y'v+py v=q v \\ \text{ \choose } v'=vp \\ \text{ so we have } (yv)'=qv \\ \text{ then integrate both sides } \\ yv=\int\limits qv dx +C \\ y=\frac{1}{v} \int\limits qv dx +\frac{C}{v}\]

Answer 28

And earlier we said v'=vp
so we have
\[\frac{dv}{dx}=vp \\ \frac{dv}{v}=p dx \\ \int\limits \frac{dv}{v} = \int\limits p dx \\ \ln|v|=\int\limits pdx+C \text{ we will using this } C=0 \\ \ln|v|=\int\limits p dx \\ v=e^{\int\limits p dx}\]

Answer 29

like this is the method using the integrating factor

Answer 30

thanks a lot!

Answer 31

where the solution is
\[y=\frac{1}{v} \int\limits qv dx+\frac{C}{v} \\ \text{ where } v=e^{\int\limits pdx}\]

Answer 32

you're brilliant! thank you!