Question

The differential representing the family of curves y^2=2c(x+√c) where c>0 is a parameter is of order and degree???
My ans is order-2 n degree-1

Answer 1

Bt it's wrong ....How?

Answer 2

what is the differential equation?

Answer 3

If u differentiate it u will get a differential equation....

Answer 4

Does this mean something like\[y^2=2c(x+\sqrt c)\]is a solution to some differential equation, and we have to find that?

Answer 5

order I recall is the order of the derivative of the highest order present in the equation is the order of the equation

Answer 6

Yep @parthkohli it's the solution which is given and we need the differential equation

Answer 7

yes, but what is the specific differential equation for the given curve

Answer 8

Oh, I can kind of see the differential equation in this because of terms like \(y^2/2\) and \(c^{3/2}\). Again, I've not studied this but let me try to come up with something, yeah.

Answer 9

Well I always differentiate the equation given untill i eliminate the arvitary constant in it...

Answer 10

Sorry arbitary*

Answer 11

Like if u differentiate the expression first time u will get
2ydy/dx=2c
Second time diff will yield
Yd2y/dx^2 +(dy/dx)^2=0

Answer 12

2ydy/dx=2c

I think they are asking about the order and degree of this equation

Answer 13

Nope , we nees to remove the arbitrary constants in order to get the differential equation

Answer 14

Well, I got this:\[y'' = -\frac{y'^2}{y}\]

Answer 15

Need*

Answer 16

Yep parth...this is correct that's what i wrote above...

Answer 17

Oh great, exactly the same.

Answer 18

From this v can see that highest order of differential is 2 n it's degree is 1

Answer 19

Yeah, I guess you're right.

Answer 20

Bt the ans is very different from what v r getting. .

Answer 21

Does the degree of a DE reflect the highest power of the variable, or can it also include the powers of the derivatives?

Answer 22

what is the answer?

Answer 23

Order -1 , degree -3

Answer 24

Okk well @parthkohli order is the highest order derivative ...n degree is the order's power....

Answer 25

Likewise in the present situation we r having y" as the highest derivative n so order wud be 2 n it's power is 1 so that turns out to be the degree of differential

Answer 26

One arbitrary constant in the general solution implies that the order of the corresponding differential equation is exactly 1.

Answer 27

Yep @rational bt here we r having two arbitary constants....

Answer 28

I see only one arbitrary constant, c ?

Answer 29

If you open the brackets u will get
Y^2=2cx + c√c
Which can be rewritten as
Y^2=2cx + k
Here c n k are two arbitrary constant

Answer 30

Okk sorry 2c√c....

Answer 31

\(k\) is not arbitrary if it depends on \(c\)

Answer 32

Okk....bt what about the differential we formed over here ....as per that we r getting order as 2...

Answer 33

Hey rational, I wonder... is there a nice way to define "order" of a differential equation? Like do we just call it the most efficient way to express a differential equation?

Answer 34

Bt c is also arbitary then if any parameter is arbitary n some other thing depends on it then it is also arbitrary isn't it?

Answer 35

Yes, but you don't have two different arbitrary constants. You get to control only one value which changes the value at both places

Answer 36

Hey @ParthKohli compare this with the difference equations

Answer 37

or recurrence relations

Answer 38

How many initial values do you need to know to solve the particular solution of the following recurrence relation ?
\[a_{n+2} = a_{n+1}+a_n\]

?

Answer 39

Yep exactly @rational i agree with you...
If we put the value of c to be something then that wud imply the change in value of k as well...

Answer 40

Two of them. I agree. But what if we're specified with a differential equation with a huge order, and we want to know if it's the best way of being expressed?

Answer 41

In a way, I cannot claim that \(x = 1\) is a better way of expressing \(x^2 = 1\) so my question is kinda stupid. Is it the same with DE?

Answer 42

Yes,
The differential equations, \(y' = 2\) and \(y''=0\) model different problems.

Answer 43

Ah, nice. Do these also model different problems?\[y'' = -y\]\[y''' = -y'\]

Answer 44

Ofcourse, they are different

Answer 45

First equation models block spring oscillator without damping

Answer 46

if i remember correctly..

Answer 47

Yep...

Answer 48

second equation may model some other situation with some damping... i don't know...

Answer 49

Hmm, but seems that the only difference between these two DEs is an added arbitrary constant. So they're not as different as, say, \(y' = 2\) and \(y'' = 0\).

Well, alright, whatever I'm saying is subjective. lol

Answer 50

Bt @rational can we differentiate the equation infinite times to get the order of the differential..
In such a way we can have any number of possibility of order ..

Answer 51

I were taught that we have to jst eliminate the arbitary constants that's it !!!!!!

Answer 52

\(y''' = -y'\) can be thought of as the following system :

\(y' = x\)
\(x''=-x\)

Answer 53

Let me think of a better reason to see them as different...

Answer 54

Nah, I totally understand now. Thanks :)

Answer 55

@samigupta8 in general, yes.

In our specific problem, we're given the number of arbitrary constants. So order is fixed...

Answer 56

Bt it is always like that we would be given with a specific problem and which would contains some arbitrary constant and the order is going to be their count....

Answer 57

Let me ask you a question

Answer 58

N @rational shall we not follow the procedure as mentioned by @parthkohli and me which involves differentiation of the given curve to get differential

Answer 59

Yep sure....

Answer 60

\[y y' = c \]Now\[y^2 = yy' x + y^{3/2} y'^{3/2}\]\[ (y^2 -yy'x)^2 = y^3 y'^3 \]Well, I'm not sure again, because all this degree stuff is pretty sensitive.

Answer 61

Solve the following initial value problem with \(y(0) = \sqrt{2}\):
\[y' = 1/y\]

Answer 62

Y^2=2x+2

Answer 63

Excellent!

Answer 64

Oh, and if you hate that \(x\) in the equation substitute something like\[\frac{y^2}{2c} -\sqrt c=x\]

Answer 65

Bt what does this prove ???

Answer 66

Well now i m getting confused over the arbitrary constant even?
@rational would u please clarify it....That how to check for the count of arbitrary constant in a particular equation

Answer 67

\[y^2=2c \left( x+\sqrt{c} \right)\]
diff.w.r.t. x
\[2y \frac{ dy }{ dx }=2c*1,c=y \frac{ dy }{ dx },\sqrt{c}=\sqrt{y \frac{ dy }{ dx }}\]
diff. eq. is
\[y^2=2y \frac{ dy }{ dx }\left( x+\sqrt{y \frac{ dy }{ dx }} \right)\]
\[y=2\frac{ dy }{ dx }\left( x+\sqrt{y \frac{ dy }{ dx }} \right)\]

Answer 68

After finding the value of c to be dy/dx why didn't u differentiate it again in order to get differential..?

Answer 69

surjithayer is right..
y=2y'(x+(yy')^1/2 ) is the differential eq...
here u can see that the order is 1. and on squaring on both sides.. u can see that the degree will be three..

Answer 70

also u can say that the order is 1 just by looking at the diff eq. as the order is equal to the no.of times we must differentiate and that is equal to the no. of. arbitary constants...here there is just "1" arbitary constant which is "c"..

Answer 71

Bt why didn't u differentiate it again to get the differential???

Answer 72

u should only do the minimum diff. required and thats equal to the no. of. arbitrary constants..

Answer 73

here only one arbitrary constant so differentiate only once.