Construct a tangent field for y'=1+x+y.
I just need someone to check if I'm on the right track. see attachment for picture

- UsukiDoll

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- UsukiDoll

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- UsukiDoll

@UnkleRhaukus

- UnkleRhaukus

what method are you using there

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- UsukiDoll

see attachment for pic. I just want to see if I'm on the right track

- UsukiDoll

ummm manual like pick any point and plug in

- UnkleRhaukus

wouldn't it be more fun to solve the DE?

- UsukiDoll

I had to get the tangent field first

- UsukiDoll

which are those lines

- UsukiDoll

is it right???

- UsukiDoll

oh and then it was like get the antiderivative which is like the linear solution of the equation

- UsukiDoll

I almost got the answer which is y = -x-2 but something went off a bit. I'll draw it on paint and get it here.

- UnkleRhaukus

what are those the blue horizontals on your graph?

- UnkleRhaukus

thats very close, but you must remmber the constant of integration

- UsukiDoll

when the slope is 0, it produces a horizontal line which is why I drew it blue

- UnkleRhaukus

ah yes w, the blue horizontals are for when the constant of integration is zero

- UsukiDoll

the answer is y = -x-2 but I'm wondering how they even got rid of the squared part or is that just individual ?!?!?!

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- UnkleRhaukus

thats not right

- UsukiDoll

oh so how does it work?

- UsukiDoll

wait is my graph in the right direction

- UsukiDoll

alright the whole question is Construct a tangent field for y' =1+x+y. Find a linear solution by inspection of the3 tangent field and check it in the equation.

- UnkleRhaukus

- UsukiDoll

I've read the book and i saw anti-derivative

- UnkleRhaukus

|dw:1378111455680:dw|

- UnkleRhaukus

|dw:1378111488816:dw|

- UnkleRhaukus

|dw:1378111517817:dw|

- UsukiDoll

:O crap you mean I did all those lines for jacks?

- UnkleRhaukus

if all the blue line were as ive drawn them, then they would fit the pattern of the other red line on tht upper right of the graph

- UsukiDoll

ok but doesn't the lines create a curve?

- UnkleRhaukus

i think ive worked the problem out

- UsukiDoll

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- UsukiDoll

where do I go from here? I know dx x and dy y have to be in their own groups.

- UnkleRhaukus

What i think you did origionally was to plot the points for when the derivative is zero
ie when y'=1+x+y=0
and then found y=-x-1, right?
so this is a line for a solution curve

- UnkleRhaukus

|dw:1378112037836:dw|

- UsukiDoll

but the answer is y = -x-2 how the heck did they get 2?

- UsukiDoll

I did the tangent field first and tehn get the antiderivative

- UsukiDoll

which is plugging any value of x and y into the equation

- UsukiDoll

the only thing I can think of is that 2 is a typo

- UsukiDoll

for getting the antiderivative

- UsukiDoll

which for the first attempt I was close. assuming that what I did was legal

- UnkleRhaukus

for the DE
\[yâ€²âˆ’y=1+x\]
is of the form
\[yâ€²+p(x)y=q(x)\]
use an integrating factor \(Î¼=e^{âˆ«p(x)dx}\)
so the equation becomes
\[(yÎ¼)â€²=q(x)Î¼\]

- UsukiDoll

:O but I havent' learned that yet... x.x.x..x.x.x.x

- UnkleRhaukus

well you can not seperate the seriables

- UsukiDoll

>:O really?

- UsukiDoll

then why the hell is the book showing anti-derivative and why is it -x-2?!?!!??!

- UnkleRhaukus

i dont know what method you are ment to used

- UsukiDoll

The boook had something called anti-derivative the whole equation and using separate variables

- UsukiDoll

@experimentX

- experimentX

huh? you are plotting the solution (integral lines) of the DE?

- UsukiDoll

really? ok but how do I prove that it's y = -x-2
When I read the book, it took the anti0derivative of the equation via separate variables

- UsukiDoll

yeah my prof is making me read an old book stupid huh

- experimentX

hmm ... you cannot separate variables here. it's a type of equation called linear equation.
have you ever heard of operator called linear operator?

- UsukiDoll

no. not that I know of.

- experimentX

the operator 'f' is called linear if
f(x+y) = f(x) + f(y)
f(ax) = af(x) for come const. 'a'

- UsukiDoll

so 1 +f(x)+f(y)?

- experimentX

no ... i mean linear operator has this property
(d/dx - 1) on the left you can extract this.

- experimentX

this thing operates on y.

- experimentX

when you encounter the problem of type
y' + g(x) y = h(x) <--- this is called first order linear equation.

- UsukiDoll

|dw:1378114795870:dw|

- experimentX

|dw:1378114858665:dw|

- experimentX

how do you differentiate
\[ y e^{p(x)} = q(x) \]

- experimentX

*y(x)

- UsukiDoll

|dw:1378115097555:dw|

- experimentX

ok ok ... that's good, how do you differentite
\[ y(x) e^{p(x)} = q(x) \]

- UsukiDoll

I want to say by separating the variables but I iknow that's not it for real. ummm ...arghhhhh I've seen that before but it was in second order differential equations

- UsukiDoll

like v(x) = e integral p(x)

- experimentX

no no ... just differentaite the above equation

- experimentX

and see what you get

- UsukiDoll

hmmmmm y= 1 +x^2/2+y^2/2

- UsukiDoll

+c cuz of constant or maybe that's not required

- experimentX

no lol ... i meant don't you get
\[ y'(x)e^{p(x)} + p'(x)e^{p(x)} = q'(x) \]

- experimentX

compare this with your equation
\[ e^{p(x)}(y'(x) - y) = e^{p(x)}(1 + x)\]

- experimentX

and find out what is p(x)

- UsukiDoll

so the p(x) is unknown

- UsukiDoll

|dw:1378115813841:dw|

- experimentX

no no ... compare it with what i said earlier ... what information do you get on p(x)

- UsukiDoll

exponential stuff

- UsukiDoll

|dw:1378116102972:dw|

- experimentX

|dw:1378116160628:dw|

- experimentX

you get the information that p'(x) is ..

- UsukiDoll

if only my book actualy had that

- experimentX

it's easy ... you can find it yourself :P

- experimentX

p(x) = something whose derivative is -1

- UsukiDoll

-___- see depends on the book because my prof is making me read a vintage one wherer most modern ones are way the heck better

- UsukiDoll

and easier to understand. x.x.x.xx

- UsukiDoll

arghhhh >.<

- experimentX

hmm ... which book are you reading?

- UsukiDoll

Differential Equations: A Concise Course (Dover Books on Mathematics) by H. S. Bear

- UsukiDoll

IT sux

- experimentX

you are engg major or math major?

- UsukiDoll

math major...currently taking two courses.

- UsukiDoll

this and calc 4

- UsukiDoll

I just wish the question say slope field...got it already...

- experimentX

Differential equations is quite hectic subject, this thing might be useful for you
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/index.htm

- UsukiDoll

I just reread a comment integrate doh.
|dw:1378117367801:dw|

- UsukiDoll

|dw:1378117488172:dw|

- experimentX

no ... p(x) is -x

- experimentX

your equation is
\[y' - y = 1 + x \]
the standard equation is
\[ e^{p(x)}y' - p'(x)ye^{p(x)} = q(x) \]

- experimentX

sorry edit
\[ e^{p(x)}y' + p'(x)ye^{p(x)} = q(x) \]

- UsukiDoll

|dw:1378117802162:dw|

- experimentX

no no ... we don't need to do that. our assumption was that DE is of the form
\[ \frac{d}{dx}(e^{p(x)}y) = e^{p(x)}q(x)\]

- experimentX

no integration ...just we had \( p'(x) = -1 \implies p(x) = -x \)

- experimentX

so we have something like
\[ \frac{d}{dx}(e^{-x} y) = e^{-x}(x+1) \]
can you integrate both sides?

- UsukiDoll

ok ummm... if I just do e^-x, the antiderivative is -e^-x

- UsukiDoll

y would be y^2/2 and x would be x^2/2

- experimentX

http://www.wolframalpha.com/input/?i=integrate+e%5E%28-x%29%28x%2B1%29

- experimentX

\[ e^{-x}y(x) = C- e^{-x}(x+2)\]

- experimentX

\[ \int d(e^{-x}y(x)) = \int e^{-x}(x+1) dx+ c\]

- experimentX

now multiply both sides with e^(x) to isolate y(x)

- UsukiDoll

|dw:1378119219209:dw|

- experimentX

No ... you have to integrate first.

- UsukiDoll

|dw:1378119465770:dw|

- experimentX

\[ \int d(e^{-x}y(x)) = \int e^{-x}(x+1) dx+ c \]

- experimentX

|dw:1378119763855:dw|

- UsukiDoll

arggghhhhhh curse the book hgjgghjgj

- UsukiDoll

|dw:1378120544268:dw|

- UsukiDoll

>:/

- UsukiDoll

I'm burned out and tired. Can we work on this tomorrow?

- UsukiDoll

Now I know what you guys are talking about. The equation has to be in the DE standard form of dy/dx+P(x)y=Q(x)
y' = 1+x+ y
I have to put it in standard form so subtract y from the right
y'-y = 1+x
so now I have to identify which is p(x) = -1 and q(X) = 1+X
now to get the v(x) it's e to the integral of p(x) since p(x) = -1 the antiderivative is -x so I got e^-x
so now I multiply the v(x) which results in
e^-x y = e^-x (1+X)
If I integrate from both sides I get the following result
-e^-x y = -e^-x (x+2)
Now how the heck do I get rid of the e^-x?

- UsukiDoll

@experimentX

- UsukiDoll

@UnkleRhaukus

- experimentX

\[ d(e^{-x} y ) = e^{-x} (1+x) dx \\
e^{-x}y = e^{-x}(x+2) +
c\]
multiply both sides by e^x

- UsukiDoll

awesomeness got it now!

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