Question

Solve the given differential equation using the proper substitution:
dy/dx=(x+y+1)^2
Someone...anyone...?

Answer 1

can i ask what is proper substitution ?

Answer 2

wolfram makes good use of \[v(x)=x+y(x)\]

Answer 3

u can do it with \(z=x+y+1\) but i dont know...is this proper substitution?

Answer 4

it starts out as u=x+y+1 and then derives u to get du=1+dy/dx. After this it becomes dx=(du/1+u^2) and I can follow the algebra; I'm just confused as to what's going on and why.

Answer 5

your changeing the function into hopefully an easier problem thru the substitution

Answer 6

change of variables is another phrase ive seen used

Answer 7

Hmm...I need to see this kind of problem solved so that I can see the process I think. I've had trouble finding one similar online. I have the solutions manual but I need to understand what's going on and there are several steps which are skipped.

Answer 8

Perhaps I should continue my search...

Answer 9

the first setup: dy/dx tells us that y is a function of x

we want to find an easier setup for us to work with and solve for

u = x + y(x) + 1 ; derive it to determine a suitable du/dx, and possibly a dy/du since:\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\]
du/dx = dx/dx + dy/dx + 0
= 1 + dy/dx

yeah, it prolly help to keep searching :)

Answer 10

Yeah, I'm on the verge of understanding here. I'm not quite sure why the solution goes on to simplify in terms of dx though. It reads dx=du/(1+u^2) and then integrates this. I'm confused as to why and what is happening here. I shall continue searching. Thank you kindly for your help though! :)

Answer 11

\[\frac{du}{dx}=1+\frac{dy}{dx}\to\ \frac{dy}{dx}=\frac{du}{dx}-1\]
using this then
\[\frac{dy}{dx}=u^2\]
\[\frac{du}{dx}-1=u^2\]
\[\frac{du}{dx}=u^2+1\]

Answer 12

Yes, this is exactly what the solution is giving me. Why are we solving for dx by itself though?

Answer 13

by separating this, we can solve for du
\[\frac{du}{u^2+1}=dx\]
\[\int \left(\frac{du}{u^2+1}=dx\right)\]
\[\int \frac{du}{u^2+1}=\int dx\]

Answer 14

Okay I see...

Answer 15

now, since we know that : u = x+y+1

once we find a suitable "u" in terms of x, we can solve for y

y = u-x-1

Answer 16

wait wait wait...why all of a sudden are they u-x-1 rather than u+x+1? Why did the signs change?

Answer 17

remember how we defined the substitution:

u = x + y + 1 ; but we dont want "u", we want "y"

what is y

Answer 18

or rather:\[u(x)=x+y(x)+1\]define \(y(x)\)
\[y(x)=u(x)-x-1\]

Answer 19

Ahhhh...so jump back a step though? Are we not integrating both sides of the equation where we separated du/(1+u^2) and dx?

Answer 20

we are

Answer 21

we are determining as solution to u(x)

Answer 22

ohhh so then I get \[\tan^{-1} (x+c)-x-1\]

Answer 23

I think I get it.

Answer 24

it looks good to me; its easier prolly to think in terms of functions in stead of variables to get your head around this

Answer 25

Yes I think you've got a point there. I just finished solving Bernoulli DEs and thought these looked easier but for some reason the substitution part threw me a curve ball. What I find kind of confusing is that the original DE had everything in the brackets squared. \[\left( x+y+1 \right)^{2}\] Where did the power or two disappear to? Was it eaten up in the substitution as well?

Answer 26

it did not disappear, we used it in the process

(x+y(x)+1)^2 ; such that u(x) = x+y(x)+1, therefore

(x+y(x)+1)^2 = \(u^2(x)\)

Answer 27

Ohhh so THAT's why we had

Answer 28

u^2 in the earlier steps. It's all making sense now. :D

Answer 29

Thank you very very much! You've been a great help!

Answer 30

youre welcome, and good luck ;)

Answer 31

Thanks, cheers!