Question

Trying to solve a straightforward DE, problem prompt below.

Answer 1

\[\text{Solve the differential equation.}\]\[\frac{dy}{dx}=2+\cos(y-x-4)\]

Answer 2

@Concentrationalizing , I'm not really sure what technique I should use here. It doesn't look separable, how do you think I should just approach this? Don't need help working through the problem itself (yet, at least)-but don't know how to solve it.

Answer 3

I don't know, I would probably just attempt to make a substitution, y-x-4=u

Answer 4

Er, dy/dx - 1

Answer 5

Yeah, youknow what I mean, haha

Answer 6

My bad for butchering that, lol.

Answer 7

Alternative possibilities would be instead of picking u, you could pick that it equals arcsin(u) or arctan(u) to cancel out the trig function with an identity if it didn't work, but I doubt those will be much use in this problem. But it can't hurt to consider it for problems down the road.

Answer 8

Alright, cool. Is this the thing/solution method related to this? To be honest I've barely studied this stuff and this is material that I did easily the worst on a test with.

Is this dealing with *homogeneous functions* of order (something)?, (NOT homogeneous DE's), or is this dealing with substitution relating to turning a nonexact DE into an exact DE? Or is this dealing with some other kind of substitution technique.

Answer 9

You can do this kind of substitution given that you have ax + by + c (something linear). It's a way of doing a separation of variables problem.

Answer 10

Well, I mean, just, name of a technique. Is this the thing where you use substitution to transform the problem into the exact differential of some mystery function mu(x) times y? Or is this something else.

Answer 11

Closely related to homogenous of n order, which is also a separation of variables technique. But I would group it all under linear first order separation of variables. Nothing to do with exactness or anything.

Answer 12

Not a specific name that im aware of.

Answer 13

Alright, separation of variables by substitution, I think I get how this is supposed to work, one sec.

Answer 14

\[du-1=2+\cos(y-x-4)\]

Answer 15

WHoops, argument of function

Answer 16

\[du-1=2+\cos(u)\]

Answer 17

What Concentrationalizing has put is not quite right. \[\Large y-x-4=u \] Differentiate both sides with respect to x. \[\Large \frac{dy}{dx}-1=\frac{du}{dx}\]

Answer 18

Wait, I think I found an example in my book to follow. This is what we're doing in concept, right? http://i.imgur.com/wjBfvkv.png

Answer 19

I was trying to give the idea, but I know I butchered it and tried to half-correct it "Er..dy/dx -1". I was hoping he knew what I meant :/

Answer 20

My book calls it, "Reduction to separation of variables", it was sort of crammed into the very end of a section on substitution.

Answer 21

Alright, gonna setup the problem like that, one sec.

Answer 22

\[\frac{du}{dx}+1=2+\cos(u)\]

Answer 23

\[\frac{du}{dx}=1+\cos(u)\]

Answer 24

So far so good.

Answer 25

(lol, that was the trivial part, one second, I just don't want to set this up wrong in the first place.)

Answer 26

Haha. Well, would be worse to have the trivial part wrong? xD

Answer 27

\[\frac{du}{dx}dx=[1+\cos(u)]dx\]

Answer 28

\[du=[1+\cos(u)]dx\]

Answer 29

\[\frac{1}{1+\cos(u)}du=dx\]

Answer 30

\[\int\limits_{}^{}\Bigg(\frac{1}{1+\cos(u)}\Bigg)du=\int\limits_{}^{}dx\]

Answer 31

(This feels wrong, am I on the right track?)

Ye wote m84

Answer 32

Its good.

Answer 33

Yeah now you have to have another realization to continue forward.

Answer 34

Identity:D

Answer 35

WE NEED TO GO DEEPER
(Lol, one sec, looking to see if I recognize the identity, I think it might be something like we have the reciprocal of a double angle formula or something)

Answer 36

I'll keep my mouth shut for this one :x

Answer 37

One sec

Answer 38

\[\cos^2(u)=\frac{1+\cos(2u)}{2};\]

Answer 39

Bingo.

Answer 40

Just make the proper adjustments of course.

Answer 41

\[\frac{1}{1+\cos(u)}=\frac{1}{2}\frac{2}{1+\cos(u)}=\frac{1}{2}\frac{1}{\cos^2(u/2)}\]

Answer 42

You definitely got it from there :)

Answer 43

Alright. Taking a shot now.

Answer 44

\[\frac{1}{2}\int\limits_{}^{}\frac{1}{\cos^2(2u)}du=x;\]

Answer 45

\[\frac{1}{2}\int\limits_{}^{}\sec^{2}(u/2)du=x\]

Answer 46

(Made a mistake in my second to last post)

Answer 47

You corrected it, though.

Answer 48

(Sorry, one sec) \[\frac{1}{2}\int\limits_{}^{}\sec^2(u/2)du=x; \]\[\int\limits_{}^{}\sec^2(u/2)[1/2(du)]=x;\]

Answer 49

\[\tan(u/2)=x\]

Answer 50

(Was that right so far?)

Answer 51

Yes. Just you know you need the + C :P

Answer 52

Alright, cool. Do I need both, or do I combine them into a single constant of integration?

Answer 53

And then, is that my solution, or how exactly do I get it in the final form of the solution?

Answer 54

*kicks internet*

And yeah, 1 constant, so tan(u/2) = x + C
But you do need to back-substitute for u and then tryto take it as far as you can to get y solved for.

Answer 55

Alright, so you're saying that...

Answer 56

(Dear internet, please forgive me) http://i.imgur.com/CQf00qB.png

Answer 57

Alright, back-substituting in one second.

Answer 58

Haha. Well...
\[\tan(u/2) = x+C\]
\[\tan(\frac{ y-x-4 }{ 2 }) = x + C\]
\[\frac{ y-x-4 }{ 2 } = \arctan(x+C)\] and then solve for y, blah blah

Answer 59

\[\tan(u/2)=x+C; \ \ \ \tan\bigg(\frac{y-x-4}{2}\bigg)=x+C.\]
Now, uh...Oh, well, you already, heh

Answer 60

Alright, yeah, this makes sense. As long as I understand the general technique, I'm going to leave it at this and move to the next one.

(The thing that confuses me about these sets of solutions is that I'm very used to the solutions in the form of c_1y_1+c_2y_2, etc, a fundamental set of solutions, each individually multiplied by arbitrary constants. )

Answer 61

Just make sure you know the substitution needs to be linear. This worked because it was u = y- x- 4. If it were u = y - x^2 - 4, wouldnt have worked.

Answer 62

Alright, thank you for mentioning that, I could have been in serious trouble if I didn't know that, heh.

Answer 63

Well, this solution was from a first order equation, one where we're pretty much always integrating and getting the typical + C in finding our solution. When we find a solution to something like y'' + 5y' + 6y = e^t, we're not even integrating. You're solving a characteristic polynomial and then doing undetermned coefficients or your method of choice, no + C from integration involved. Those are the solutions that have the c1, c2, etc.

Answer 64

Alright, cool. I just posted another ODE problem, I'm not sure I need help in it at all, but I'm going to tag you in it and start trying to work on it in case you want to help. Thanks very much, man.