Finding the degree of a deferential equation

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In summary, Cindrilla is trying to solve a problem where the degree and order of an equation are both 1, but she is getting lost. She is confused about how to proceed and needs someone to help her.
  • #1
cindrilla
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hi frinds I am new to this forum pls guide me if I am wrong at any place

im very much confused abt finding the degree of deferential equation of various problems.I have got the defination of "degree "of deferential equation as "the highest exponent of the highest derivative "okk now I am clear but actually working with problems I am much wired can anyone help me pls
 
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  • #2
Will try to help, what do you want to know?
 
  • #3
i want to to know the degree of [d^2y/dx^2+(dy/dx)^3]^6/5=6y
 
  • #4
On the left hand side (LHS) of the equals sign, is that whole term divided by 5, or is that raised to the 6/5 power?
 
  • #5
no yaar the whole of the L.H.S IS raisaed to the power of 6/5 got it?
 
  • #6
Ok... to the 6/5 power...

I would raise both LHS and RHS to the fifth power... this would cancel out the fractional power on the LHS. On the RHS we would now have (6y)^5.

On the LHS, the differential terms in brackets will now be raised to the sixth power.
If you now expand the LHS... we will find that the highest degree of the highest derivative will be [(d2^y)/(dx^2)]^6.
 
  • #7
[(LHS term)^6/5]^5 = (LHS)^6 (The fractional power cancels out)
Then expand LHS...
 
  • #8
How are you getting on with that Cindrilla? :)
Any luck?
 
  • #9
hey i too had the same idea but in vain the answer to the above question is tht it's degree is 2
 
  • #10
sorry it's degree is 1 and order is 2
 
  • #11
the power of the total l.h.s has ben shifted to r.h.s such tht (6y)^5/6so we r left with only l.h.s as no exponent so dergree is 1 order is 2 but my question is y we should do this
 
  • #12
Ok... we have raised both LHS and RHS to the fifth power...
On LHS we ought to now have this:
[d^2y/dx^2+(dy/dx)^3]^6
On RHS: (6y)^5

Do you agree with that so far?
 
  • #13
ya i agree with wht my book & u r saying but y shouldn't the problem be solved in the way how I am saying
 
  • #14
We don't have to worry about the power which dy/dx on the LHS, and y on the RHS, are raised to, we have to look at the power which d^2y/dx^2 is raised to, that is the highest order derivative... do you agree with that comment?
 
  • #15
Cindrilla, I would need to see your working out to be able to pinpoint where you may be going wrong...
can give me some idea of how you are attempting to solve this yourself please?
 
  • #16
ya first i had taken out the power 5 on both sides so tht [L.H.S]^6=(6Y)^5 THEN naturally we have to expand however after expansion we will get tht the highest exponent of d^2y/dx^2 as 6 isn't it?then i would say tht the degree is 6 & order is 2
then y is this method wrong??
 
  • #17
Cindrilla, that is correct.

I read what you posted a little earlier:
"the power of the total l.h.s has ben shifted to r.h.s such tht (6y)^5/6so we r left with only l.h.s as no exponent so dergree is 1 order is 2 but my question is y we should do this"

The problem you pose invites us to get rid of the fractional 1/5 power only.
 
  • #18
I would say that raising both sides to the 5/6 power is mathematically correct, but then if we had to solve further for y, with the term on the RHS raised to the 5/6 power, it would be much harder to solve than just getting rid of the fractional power only...
 
  • #19
But take the 5/6 power of both sides of the equation changes the equation and changes the degree of the equation.

If we have
[tex]\left(\frac{d^2y}{dx^2}\right)^4= y[/itex]
that is a fourth degree equation. To solve it, I would probably start by taking the fourth root of each side:
[tex]\frac{d^2y}{dt^2}= \pm y^{1/4}[/itex]
Those equations will have the same solutions but they are different equations. The fact that there is now no fourth power of the derivative does not change the fact that the orginal question is of degree 4.

the problem given [itex][d^2y/dx^2+(dy/dx)^3]^6/5=6y[/itex], if multiplied out, would have the second derivative to the 6/5 power so, if you want to use the term "degree" here it would be 6/5.

(I say "if you want to use the term "degree" here" because "degree" is normally used only for integer powers.)
 
  • #20
oh my frinds ur not getting my point at all okk now listen to my problem carefully
when we had the given question lyk this stated below as



{ (d^2y/dx^2)+(dy/dx)^3}^6/5=6y
then
we would naturally take first 5th power on both sides so tht the 1/5th power of l.h.s will be canceled to give (l.h.s)^6=(6y)^5
okkk
it is clear?
now if we say tht the degree here is 6 then the answr is wrong but yy?
wht is the mistake i want to to know it



ya i know tht we can take 5/6 th powr on both sides so tht l.h.s part will have no exponent whras r.h.s wil have (6y)^5/6 okkk
this is correct as given in my book
 

FAQ: Finding the degree of a deferential equation

What is a differential equation?

A differential equation is a mathematical equation that relates the rates of change of a function to the function itself. It involves one or more derivatives of the unknown function, and is used to model real-world phenomena in various fields such as physics, engineering, and economics.

What is the degree of a differential equation?

The degree of a differential equation is the highest power of the derivative appearing in the equation. For example, a first-degree differential equation would contain first-order derivatives, while a second-degree differential equation would contain second-order derivatives.

How do you find the degree of a differential equation?

To find the degree of a differential equation, you must first rearrange the equation so that the highest-order derivative is on one side and all other terms are on the other side. The power of this derivative will then be the degree of the equation.

Why is it important to know the degree of a differential equation?

The degree of a differential equation is important because it determines the type of solution that the equation will have. Higher degree equations often have more complex solutions, making them more difficult to solve. Additionally, the degree can provide insight into the behavior of the solution and the system being modeled.

Can the degree of a differential equation change?

Yes, the degree of a differential equation can change depending on the method used to solve it. For example, certain transformations or substitutions can reduce the degree of the equation, making it easier to solve. However, the degree will always remain the same in its original form.

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