# degree of differential equation

• Dec 6th 2011, 09:48 AM
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]
• Dec 6th 2011, 10:25 AM
HallsofIvy
Re: degree of differential equation
Yes, that is a first order differential equation.

Even in the first form, $e^{dy/dx}= 2$ 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.
• Dec 6th 2011, 05:32 PM
Re: degree of differential equation
Quote:

Originally Posted by HallsofIvy
Yes, that is a first order differential equation.

Even in the first form, $e^{dy/dx}= 2$ 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
• Mar 13th 2012, 07:50 AM