# Math Help - order and degree

1. ## order and degree

OKay, this is really a trivial doubt. However i need a clarification.
Find the order and degree of the following:

a) y'' + (1 + (y')^3 )^(1/2) = 0 where y' is the first derivative and y'' is the second derivative of y
Order is 2 and degree is 2. Am i right?

b) y'' + (1 + (y')^2)^(1/2) = 0
Here again order is 2 and degre is 2. am i right?

2. Originally Posted by poorna
OKay, this is really a trivial doubt. However i need a clarification.
Find the order and degree of the following:

a) y'' + (1 + (y')^3 )^(1/2) = 0 where y' is the first derivative and y'' is the second derivative of y
Order is 2 and degree is 2. Am i right?

b) y'' + (1 + (y')^2)^(1/2) = 0
Here again order is 2 and degre is 2. am i right?
Hello poorna,

The order is the highest derivative in the equation and the degree is the power of this derivative after the equation has been put into rational form and cleared of fractions. So the answer is -no- you are not correct. The orders are indeed 2 but the degree is in both cases ...

Let's give a more complicated example:

$\left(y"\right)^3+y"\cdot \left(\left(y'\right)^2+1\right)^3=0$

Here we have order 2 (highes derivative is y") and degree 2 (the power is 2, don't forget to eliminate the common y").

Do you see what your degrees are now?

Coomast

3. @ Coomast
I understood the example you gave.
However in the questions I asked, I am still not quit sure why the degree is not 2.
y'' + (1 + (y')^3 )^(1/2) = 0
=> y'' = -(1 + (y')^3 )^(1/2)
Here there is a (1/2) in the power of the term in teh RHS. DOnt i have to remove it by squaring, and so teh power of y'' becomes 2? Am I making sense. I still dont know.

4. Originally Posted by poorna
@ Coomast
I understood the example you gave.
However in the questions I asked, I am still not quit sure why the degree is not 2.
y'' + (1 + (y')^3 )^(1/2) = 0
=> y'' = -(1 + (y')^3 )^(1/2)
Here there is a (1/2) in the power of the term in teh RHS. DOnt i have to remove it by squaring, and so teh power of y'' becomes 2? Am I making sense. I still dont know.
The order is 2 (the highest derivative is y'') and the degree is 1. You must not take the square in order to eliminate the root, it has nothing to do with the y''. OK, let's give another example:

$y''=1+(y')^2$

The order is 2 and the degree is 1. Another example:

$(y'')^{\frac{1}{2}}=1+(y')^2$

Order 2 and degree 1, because here you need to make the power a number. Is this a bit clear? If not do come back.

Coomast

5. Oh yes now its fine thanks a lot