Question

x'+3x=2, x(0)=1

Answer 1

you can rearrange for x' right?

Answer 2

I don't see why not

Answer 3

so x' = 2-3x rigth?
what do you have to do next?

Answer 4

I need to know what x is

Answer 5

x'+3x=2
what if we multiplied both sides of the equation by a v
x'v+3xv=2v

but we chose this v such v'=3v
we can solve v'=3v for v using separation of variables

so we have
\[\frac{dv}{dx}=3v\]
\[\frac{1}{3v} dv= dx\]
integrate both sides
\[\frac{1}{3}lnv=x+K\]
let this constant K be zero

\[v=e^{3x}\]

so we have
\[x'e^{3x}+3xe^{3x}=2e^{3x}\]
reverse the product rule we can now write
\[(xe^{3x})'=2e^{3x}\]
now integrate both sies
\[xe^{3x}=2*\frac{1}{3}e^{3x}+C\]
wait a minute i think i just over complicated things lol

Answer 6

let me turn my air condition on

Answer 7

its a little confusing because we are using x as the dependent variable. You should have a t in your e^3x

Answer 8

yep you are right

Answer 9

\[\frac{dv}{dt}=3v\]

Answer 10

wow i think i had no idea

Answer 11

x is a function of t here. so your v should be
\[v = e^{3t}\]
when you multiply everything out you get:
\[x'e^{3t}+3e^{3t}x = 2e^{3t}\]
\[(xe^{3t})' = 2e^{3t}\]

Answer 12

\[\frac{1}{3v} dv =dt\]

Answer 13

The answer to this ODE is \[x(t) = 1/3*(\exp(-3t)+2) \]
I just need the steps to get there...

Answer 14

\[\frac{1}{3}lnv=t\]

Answer 15

\[v=e^{3t}\]

Answer 16

\[(xe^{3t})'=2e^{3t}\]
integrate both sides
\[xe^{3t}=2*\frac{1}{3}e^{3t}+C\]

Answer 17

\[x=\frac{2}{3}\frac{e^{3t}}{e^{3t}}+\frac{C}{e^{3t}}=\frac{2}{3}+e^{-3t}C\]

Answer 18

now we also have an initial condition
x(0)=1

Answer 19

\[1=\frac{2}{3}+e^{-3*0}C=\frac{2}{3}+C => C=1-\frac{2}{3}=\frac{1}{3}\]

Answer 20

so we have
\[x=\frac{2}{3}+\frac{1}{3}e^{-3t}=\frac{1}{3}(2+e^{-3t})\]

Answer 21

While I understand what you're doing in the latter part of the problem, I'm a little confused with the v notation... not to be picky, but is there a way to explain the initial setup without it?

Answer 22

there is an actual formula
i can derive it for you if you like so you not have to go through the long process i just like doing it because it explains why we get the solution we do

Answer 23

one sec k?

Answer 24

sure, I enjoy derivations!

Answer 25

pretend we have this general linear equation:
\[y'+p(x)y=q(x)\]
now if we choose to multiply both sides by v(x) does the equation still hold?
this is the part where you say yes lol
\[v(x)y'+v(x)p(x)y=v(x)q(x)\]
now why would we multiply by v(x)
we can choose v(x) to be anything
what is the product rule (uv)'=uv'+u'v right?
so vy'+v'y=(yv)'
correct?

so what if we choose a v such that v'(x)=v(x)p(x)

so we have
\[\frac{dv}{dx}=vp\]
now we can use separation of variables to solve for v
\[\frac{1}{v}dx=p dx\]
now integrate both sides
\[lnv=\int\limits_{}^{}p dx\]
so we always want to choose a v such that v>0 \[e^{lnv}=e^{\int\limits_{}^{} p dx}\]
\[v=e ^{\int\limits_{}^{}p dx}\]
so we have
\[e^{\int\limits_{}^{}p(x) dx}y'(x)+e^{\int\limits_{}^{}p(x) dx}p(x)y(x)=e^{\int\limits_{}^{}p(x) dx}q(x)\]
good so far?

Answer 26

lets just verify a couple of things:
derivative of y is y'
derivative of \[e^{\int\limits_{}^{}p(x) dx}\]
is \[(\int\limits_{}^{} p(x) dx)'e^{\int\limits_{}^{}p(x) dx}=p(x)e^{\int\limits_{}^{}p(x) dx}\]
so we can use product rule to rewrite the left hand side

Answer 27

ahhh, v is your integrating factor! I'm glad you explained where it actually came from; I know the form e^P(x) but didn't know where or why it came from... I appreciate your help!

Answer 28

\[(ye^{\int\limits_{}^{}p(x)dx})'=q(x)e^{\int\limits_{}^{}p(x) dx}\]

Answer 29

now integrating both sides we can see we can obtain y

Answer 30

\[ye^{\int\limits_{}^{}p(x) dx}=\int\limits_{}^{}q(x)e^{\int\limits_{}^{} p(x) dx} dx+C\]

Answer 31

\[y=\frac{1}{e^{\int\limits_{}^{}p(x) dx}}\int\limits_{}^{}q(x)e^{\int\limits_{}^{}p(x) dx} dx+\frac{C}{e^{\int\limits_{}^{}p(x) dx}}\]

Answer 32

i divided both sides by \[e^{\int\limits_{}^{}p(x) dx}\]
to get y by itself

Answer 33

I think I get it, but what about if you have 3 pieces now with factors of x? see my other posted problem?