degree of differential equation

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December 6th 2011, 10:48 AM
ayushdadhwal

degree of differential equation

what can be degree of differential equation $e^{\frac{dy}{dx}}=2$
my working
if I expand $e^t$ where $t={\frac{dy}{dx}}$ then degree seems to be not defined as $t$ goes up to infinity but if i take ln(natural log)on both sides then degree seems to be 1
as we get ${\frac{dy}{dx}}=ln2$ where is the problem ? is modification changed the equation in other form[/quote]
December 6th 2011, 11:25 AM
HallsofIvy

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:
$1+ \frac{dy}{dx}+ \frac{1}{2!}\left(\frac{dy}{dx}\right)^2+ \frac{1}{3!}\left(\frac{dy}{dx}\right)^3+ \cdot\cdot\cdot$
You have only the first derivative! The powers of that first derivative give the degree not the order- it is still first order.
December 6th 2011, 06:32 PM
ayushdadhwal

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:
$1+ \frac{dy}{dx}+ \frac{1}{2!}\left(\frac{dy}{dx}\right)^2+ \frac{1}{3!}\left(\frac{dy}{dx}\right)^3+ \cdot\cdot\cdot$
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
March 13th 2012, 08:50 AM
ayushdadhwal

Re: degree of differential equation

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