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]

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.

Quote:

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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.

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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

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