Question

ODE clarification, Canonical form:

Given \(2y'+siny'=x+y^2~~~and~~~y(0)=0\) Find the canonical form.

Answer 1

So I got \[y''=\frac{1+2yy'}{2+cosy'}\] as the canonical form. What am I supposed to do with the y(0)=0? My teacher put "\(2y'(0)+siny'(0)=0+y^2(0)\Rightarrow y'(0)=0\)" on the board

Answer 2

@vishweshshrimali5 , any good with this, It's not in my book, so I am jsut not sure on the procedure

Answer 3

I just don't see how he solved that for y', like clearly zero works, but is it the only possible solution?

Answer 4

Basically he used y'(0) for finding out the exact value of the integration constant C

Answer 5

there is no integration constant

Answer 6

no integration going on here

Answer 7

See, whenever you solve a differential equation you basically just integrate it. This integration results in integration constant.

A simple differential equation has infinite solutions but one with an initial value has only one solution - because C has been specified

Answer 8

but this is not solving, it is canonical form

Answer 9

Canonical form is prior to any integration

Answer 10

I am not much familiar with canonical form... sorry :(

But, @Kainui and @ganeshie8 can help you out...

Again sorry

Answer 11

it's cool, I'm just wondering if it was a guess, the real issue is the algebra here: \(2y′(0)+siny′(0)=0+y2(0)⇒y′(0)=0\) He had to solve for y'(0), but how do you solve 2y'+siny'=0?

Answer 12

non-linear odes are not nice

Answer 13

u have to linearize this first

Answer 14

No, you don't. I'm not solving it. It's just the canonical form

Answer 15

ie get the highest derivative by itself on the left, and that's it. He added this bound like thing and never explained it

Answer 16

@Kainui

Answer 17

if you want my work for canonical, I can type that up

Answer 18

Yeah that's exactly what I was gonna ask for haha. Show me your work and we'll see what we can figure out from there, I have some ideas.

Answer 19

ok so like I said before, my final answer is up top, but her is the work for that:
\[2y'+siny'=x+y^2\]\[[2y'+siny']'=[x+y^2]'\]\[2y''+y''cosy'=1+2yy'\]\[y''[2+cosy']=1+2yy'\]\[y''=\frac{1+2yy'}{2+cosy'}\]

Answer 20

now he did this problem on the board, he then wrote what I quoted above and said the final answer was what I have + y(0)=0 and y'(0)=0

Answer 21

So it was given that y(0)=0 but then showed that y'(0)=0 too? Or was that also given

Answer 22

but the issue I really am having is how (other than guessing) he determined that y'(0)=0

Answer 23

nope, showed that y'(0)=0

Answer 24

he just plugged and chugged into the original given eq

Answer 25

This isn't in the book, so I have no reference

Answer 26

The way I did it is weird, so I'm trying to find a better method one sec.

Answer 27

ok

Answer 28

Oh I see! Haha it's kinda fun. Ok check it out, take the original equation and plug in y(0)=0

Answer 29

right, I get 2y'+siny'=0

Answer 30

but I'm missing why x=0 is given

Answer 31

So now solve for y' in terms of itself we get: \[\Large y'(0)=- \frac{1}{2} \sin( y'(0))\]
Here we can ask ourselves, sine of what number times a constant is equal to itself?

Answer 32

ohhhhh nice

Answer 33

ok that makes sense now

Answer 34

You can draw this as a graph if it is not clear and it will make it immediately apparent, look where y=x and y=-1/2sin(x) intersect.

Answer 35

right right, you're right

Answer 36

ok, he just didn't explain that reasoning on a new topic....

Answer 37

what's the phrase, couldn't see the trees for the forest? I think

Answer 38

Thanks

Answer 39

Yeah the original way I did it was ridiculous, I just took this limit by plugging itself in infinitely to itself haha. \[\Large y' = - \frac{1}{2} \sin ( - \frac{1}{2} \sin ( - \frac{1}{2} \sin (...)))\]

Answer 40

yea, that may be a bit much lol

Answer 41

But since we have the easy way I just thought for laughs I'd share XD

Answer 42

:) thanks laughs are always a good thing.