Thread: order and degree

(d2y/dx2) + (dy/dx)^0.5 +y =0

In this equation is the degree 2 or cannot be said?
since after removing the fraction there will be both degree one and degree 2
of the derivative (d2y/dx2)
or do we only take the highest degree of the highest order?

Originally Posted by Dili

(d2y/dx2) + (dy/dx)^0.5 +y =0

In this equation is the degree 2 or cannot be said?
since after removing the fraction there will be both degree one and degree 2
of the derivative (d2y/dx2)
or do we only take the highest degree of the highest order?

We take the degree of the highest order derivative.

But note, since the first derivative is to a power, this is a non-linear equation. So it is a "second order non-linear ordinary differential equation."

-Dan

I still don't get it

Our lecturer said we first have to remove fractions and then find the degree
then,

(d2y/dx2)^2 +2y(d2y/dx2) + y^2 = dy?dx

so we have two degrees of d2y/dx2.
Do we take the highest which is 2 or we cannot say the degree exactly?

Originally Posted by Dili

I still don't get it

Our lecturer said we first have to remove fractions and then find the degree
then,

(d2y/dx2)^2 +2y(d2y/dx2) + y^2 = dy?dx

so we have two degrees of d2y/dx2.
Do we take the highest which is 2 or we cannot say the degree exactly?

All right, it works out to be the same in the end. The degree of the differential equation is the highest number of derivatives that we take of the unknown function (in this case y), not the highest power as it would be if this were an algebraic equation. So the degree still ends up as being 2. (And my description of the type of equation is still accurate.)

-Dan

Here is an easy criterion for a differencial equation (ordinary or partial) to be linear.

Let be two solutions.

Then is a solution also? Where are any real numbers.

If yes, then the equation is linear, otherwise it is not.