UsukiDoll
  • UsukiDoll
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
Mathematics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
UsukiDoll
  • UsukiDoll
1 Attachment
UsukiDoll
  • UsukiDoll
@UnkleRhaukus
UnkleRhaukus
  • UnkleRhaukus
what method are you using there

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UsukiDoll
  • UsukiDoll
see attachment for pic. I just want to see if I'm on the right track
UsukiDoll
  • UsukiDoll
ummm manual like pick any point and plug in
UnkleRhaukus
  • UnkleRhaukus
wouldn't it be more fun to solve the DE?
UsukiDoll
  • UsukiDoll
I had to get the tangent field first
UsukiDoll
  • UsukiDoll
which are those lines
UsukiDoll
  • UsukiDoll
is it right???
UsukiDoll
  • UsukiDoll
oh and then it was like get the antiderivative which is like the linear solution of the equation
UsukiDoll
  • 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
  • UnkleRhaukus
what are those the blue horizontals on your graph?
UnkleRhaukus
  • UnkleRhaukus
thats very close, but you must remmber the constant of integration
UsukiDoll
  • UsukiDoll
when the slope is 0, it produces a horizontal line which is why I drew it blue
UnkleRhaukus
  • UnkleRhaukus
ah yes w, the blue horizontals are for when the constant of integration is zero
UsukiDoll
  • 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 ?!?!?!
1 Attachment
UnkleRhaukus
  • UnkleRhaukus
thats not right
UsukiDoll
  • UsukiDoll
oh so how does it work?
UsukiDoll
  • UsukiDoll
wait is my graph in the right direction
UsukiDoll
  • 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.
UsukiDoll
  • UsukiDoll
I've read the book and i saw anti-derivative
UnkleRhaukus
  • UnkleRhaukus
|dw:1378111455680:dw|
UnkleRhaukus
  • UnkleRhaukus
|dw:1378111488816:dw|
UnkleRhaukus
  • UnkleRhaukus
|dw:1378111517817:dw|
UsukiDoll
  • UsukiDoll
:O crap you mean I did all those lines for jacks?
UnkleRhaukus
  • 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
  • UsukiDoll
ok but doesn't the lines create a curve?
UnkleRhaukus
  • UnkleRhaukus
i think ive worked the problem out
UsukiDoll
  • UsukiDoll
1 Attachment
UsukiDoll
  • UsukiDoll
where do I go from here? I know dx x and dy y have to be in their own groups.
UnkleRhaukus
  • 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
  • UnkleRhaukus
|dw:1378112037836:dw|
UsukiDoll
  • UsukiDoll
but the answer is y = -x-2 how the heck did they get 2?
UsukiDoll
  • UsukiDoll
I did the tangent field first and tehn get the antiderivative
UsukiDoll
  • UsukiDoll
which is plugging any value of x and y into the equation
UsukiDoll
  • UsukiDoll
the only thing I can think of is that 2 is a typo
UsukiDoll
  • UsukiDoll
for getting the antiderivative
UsukiDoll
  • UsukiDoll
which for the first attempt I was close. assuming that what I did was legal
UnkleRhaukus
  • UnkleRhaukus
for the DE \[y′−y=1+x\] is of the form \[y′+p(x)y=q(x)\] use an integrating factor \(μ=e^{∫p(x)dx}\) so the equation becomes \[(yμ)′=q(x)μ\]
UsukiDoll
  • UsukiDoll
:O but I havent' learned that yet... x.x.x..x.x.x.x
UnkleRhaukus
  • UnkleRhaukus
well you can not seperate the seriables
UsukiDoll
  • UsukiDoll
>:O really?
UsukiDoll
  • UsukiDoll
then why the hell is the book showing anti-derivative and why is it -x-2?!?!!??!
UnkleRhaukus
  • UnkleRhaukus
i dont know what method you are ment to used
UsukiDoll
  • UsukiDoll
The boook had something called anti-derivative the whole equation and using separate variables
UsukiDoll
  • UsukiDoll
@experimentX
experimentX
  • experimentX
huh? you are plotting the solution (integral lines) of the DE?
UsukiDoll
  • 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
  • UsukiDoll
yeah my prof is making me read an old book stupid huh
experimentX
  • 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
  • UsukiDoll
no. not that I know of.
experimentX
  • experimentX
the operator 'f' is called linear if f(x+y) = f(x) + f(y) f(ax) = af(x) for come const. 'a'
UsukiDoll
  • UsukiDoll
so 1 +f(x)+f(y)?
experimentX
  • experimentX
no ... i mean linear operator has this property (d/dx - 1) on the left you can extract this.
experimentX
  • experimentX
this thing operates on y.
experimentX
  • experimentX
when you encounter the problem of type y' + g(x) y = h(x) <--- this is called first order linear equation.
UsukiDoll
  • UsukiDoll
|dw:1378114795870:dw|
experimentX
  • experimentX
|dw:1378114858665:dw|
experimentX
  • experimentX
how do you differentiate \[ y e^{p(x)} = q(x) \]
experimentX
  • experimentX
*y(x)
UsukiDoll
  • UsukiDoll
|dw:1378115097555:dw|
experimentX
  • experimentX
ok ok ... that's good, how do you differentite \[ y(x) e^{p(x)} = q(x) \]
UsukiDoll
  • 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
  • UsukiDoll
like v(x) = e integral p(x)
experimentX
  • experimentX
no no ... just differentaite the above equation
experimentX
  • experimentX
and see what you get
UsukiDoll
  • UsukiDoll
hmmmmm y= 1 +x^2/2+y^2/2
UsukiDoll
  • UsukiDoll
+c cuz of constant or maybe that's not required
experimentX
  • experimentX
no lol ... i meant don't you get \[ y'(x)e^{p(x)} + p'(x)e^{p(x)} = q'(x) \]
experimentX
  • experimentX
compare this with your equation \[ e^{p(x)}(y'(x) - y) = e^{p(x)}(1 + x)\]
experimentX
  • experimentX
and find out what is p(x)
UsukiDoll
  • UsukiDoll
so the p(x) is unknown
UsukiDoll
  • UsukiDoll
|dw:1378115813841:dw|
experimentX
  • experimentX
no no ... compare it with what i said earlier ... what information do you get on p(x)
UsukiDoll
  • UsukiDoll
exponential stuff
UsukiDoll
  • UsukiDoll
|dw:1378116102972:dw|
experimentX
  • experimentX
|dw:1378116160628:dw|
experimentX
  • experimentX
you get the information that p'(x) is ..
UsukiDoll
  • UsukiDoll
if only my book actualy had that
experimentX
  • experimentX
it's easy ... you can find it yourself :P
experimentX
  • experimentX
p(x) = something whose derivative is -1
UsukiDoll
  • 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
  • UsukiDoll
and easier to understand. x.x.x.xx
UsukiDoll
  • UsukiDoll
arghhhh >.<
experimentX
  • experimentX
hmm ... which book are you reading?
UsukiDoll
  • UsukiDoll
Differential Equations: A Concise Course (Dover Books on Mathematics) by H. S. Bear
UsukiDoll
  • UsukiDoll
IT sux
experimentX
  • experimentX
you are engg major or math major?
UsukiDoll
  • UsukiDoll
math major...currently taking two courses.
UsukiDoll
  • UsukiDoll
this and calc 4
UsukiDoll
  • UsukiDoll
I just wish the question say slope field...got it already...
experimentX
  • 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
  • UsukiDoll
I just reread a comment integrate doh. |dw:1378117367801:dw|
UsukiDoll
  • UsukiDoll
|dw:1378117488172:dw|
experimentX
  • experimentX
no ... p(x) is -x
experimentX
  • experimentX
your equation is \[y' - y = 1 + x \] the standard equation is \[ e^{p(x)}y' - p'(x)ye^{p(x)} = q(x) \]
experimentX
  • experimentX
sorry edit \[ e^{p(x)}y' + p'(x)ye^{p(x)} = q(x) \]
UsukiDoll
  • UsukiDoll
|dw:1378117802162:dw|
experimentX
  • 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
  • experimentX
no integration ...just we had \( p'(x) = -1 \implies p(x) = -x \)
experimentX
  • experimentX
so we have something like \[ \frac{d}{dx}(e^{-x} y) = e^{-x}(x+1) \] can you integrate both sides?
UsukiDoll
  • UsukiDoll
ok ummm... if I just do e^-x, the antiderivative is -e^-x
UsukiDoll
  • UsukiDoll
y would be y^2/2 and x would be x^2/2
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=integrate+e%5E%28-x%29%28x%2B1%29
experimentX
  • experimentX
\[ e^{-x}y(x) = C- e^{-x}(x+2)\]
experimentX
  • experimentX
\[ \int d(e^{-x}y(x)) = \int e^{-x}(x+1) dx+ c\]
experimentX
  • experimentX
now multiply both sides with e^(x) to isolate y(x)
UsukiDoll
  • UsukiDoll
|dw:1378119219209:dw|
experimentX
  • experimentX
No ... you have to integrate first.
UsukiDoll
  • UsukiDoll
|dw:1378119465770:dw|
experimentX
  • experimentX
\[ \int d(e^{-x}y(x)) = \int e^{-x}(x+1) dx+ c \]
experimentX
  • experimentX
|dw:1378119763855:dw|
UsukiDoll
  • UsukiDoll
arggghhhhhh curse the book hgjgghjgj
UsukiDoll
  • UsukiDoll
|dw:1378120544268:dw|
UsukiDoll
  • UsukiDoll
>:/
UsukiDoll
  • UsukiDoll
I'm burned out and tired. Can we work on this tomorrow?
UsukiDoll
  • 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
  • UsukiDoll
@experimentX
UsukiDoll
  • UsukiDoll
@UnkleRhaukus
experimentX
  • 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
  • UsukiDoll
awesomeness got it now!

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