degree of differential equation

what can be degree of differential equation $\displaystyle e^{\frac{dy}{dx}}=2$

my working

if I expand $\displaystyle e^t$where $\displaystyle t={\frac{dy}{dx}}$ then degree seems to be not defined as $\displaystyle t$goes up to infinity but if i take ln(natural log)on both sides then degree seems to be 1

as we get $\displaystyle {\frac{dy}{dx}}=ln2$ where is the problem ? is modification changed the equation in other form[/quote]

Re: degree of differential equation

Yes, that is a first order differential equation.

Even in the first form, [itex]e^{dy/dx}= 2[/itex] there is only a first order derivative.

**Even** if you expand in a power series:

$\displaystyle 1+ \frac{dy}{dx}+ \frac{1}{2!}\left(\frac{dy}{dx}\right)^2+ \frac{1}{3!}\left(\frac{dy}{dx}\right)^3+ \cdot\cdot\cdot$

You have only the first derivative! The powers of that first derivative give the **degree** not the order- it is still first order.

Re: degree of differential equation

Quote:

Originally Posted by

**HallsofIvy** Yes, that is a first order differential equation.

Even in the first form, [itex]e^{dy/dx}= 2[/itex] there is only a first order derivative.

**Even** if you expand in a power series:

$\displaystyle 1+ \frac{dy}{dx}+ \frac{1}{2!}\left(\frac{dy}{dx}\right)^2+ \frac{1}{3!}\left(\frac{dy}{dx}\right)^3+ \cdot\cdot\cdot$

You have only the first derivative! The powers of that first derivative give the **degree** not the order- it is still first order.

sir I am taking about degree according to our book degree and order are two different terminology.sir see this General Terms of Ordinary Differential Equations

Re: degree of differential equation

ir i am confused in order and degree as given in link above .please clear my doubt