(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?

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- Jul 31st 2007, 06:02 AMDiliorder 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? - Jul 31st 2007, 06:03 AMtopsquark
- Jul 31st 2007, 06:09 AMDili
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? - Jul 31st 2007, 06:14 AMtopsquark
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 - Jul 31st 2007, 07:29 AMThePerfectHacker
Here is an easy criterion for a differencial equation (ordinary or partial) to be linear.

Let $\displaystyle f_1\mbox{ and }f_2$ be two solutions.

Then is $\displaystyle c_1f_1+c_2f_2$ a solution also? Where $\displaystyle c_1,c_2$ are any real numbers.

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