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derivatives of second order : For example, the second order partial derivatives of a scalar function of n variables can ..... More from Wikipedia
Derivatives of Second Order
The order of a differential equation is equal to the highest derivative in the equation. The single-quote indicates differentiation. So x ' is a first derivative, while x '' is a second derivative. x ' = 1/ x is first-order..
Derivatives of Second Order
Similarly, higher order derivatives can also be define..
Similarly, higher order derivatives can also be define.. tell me if any errors www.geocities.com 27/01/09 been busy so old stuff but hey its revision for me
Learn how you can use Taylor series to derive finite difference formulas for the second derivative of a function.
Question : (a)
Obtain all the first and second derivatives of V , where
V = (1/3)y^3+(1/2)x^2- 2y^2+ y- x- xy
and find the location and nature of the stationary points.
b) A second order ordinary differential equation is given by
d y/dt - 4 dy/dt + 4y = 0,
Find the complementary function and particular integral of the differential equation
and hence write down the full general solution.
Answer : I've already answered part a) of this and you also got an answer to part b). If you could not understand either then I suggest you take a different course... More from Yahoo Answers
Answer : I've already answered part a) of this and you also got an answer to part b). If you could not understand either then I suggest you take a different course... More from Yahoo Answers
Question : (a)
Obtain all the first and second partial derivatives of V , where
V = (1/3)y^2+(1/2)x^2- 2y^2+ y- x- xy
and find the location and nature of the stationary points.
(b) A second order ordinary differential equation is given by
y/ t - 4 y/ t + 4y = 5 cos t
Find the complementary function and particular integral of the differential equation
and hence write down the full general solution.
Answer : for b) you note the operator is D - 4 D + 4 with roots 2 & 2, thus y_c = c1 e^2t + c_2 t e^2t. For y_p I will use Heaviside's techniqes....replace D by -1 {since cos t comes from roots i}...thus [D - 4D + 4] y becomes [ -4 D + 3] y = 5 cos t, now operate on both sides with [ 4D +3] to get [ 9 - 16 D ] y = [ 4D +3] { 5 cos t} = {-20 sin t + 15 cost}, but [9-16 D ] becomes [25] {D = -1}----> y_p = [-4/5] sin t +[3/5] cos t...interesting isn't it?.. More from Yahoo Answers
Answer : for b) you note the operator is D - 4 D + 4 with roots 2 & 2, thus y_c = c1 e^2t + c_2 t e^2t. For y_p I will use Heaviside's techniqes....replace D by -1 {since cos t comes from roots i}...thus [D - 4D + 4] y becomes [ -4 D + 3] y = 5 cos t, now operate on both sides with [ 4D +3] to get [ 9 - 16 D ] y = [ 4D +3] { 5 cos t} = {-20 sin t + 15 cost}, but [9-16 D ] becomes [25] {D = -1}----> y_p = [-4/5] sin t +[3/5] cos t...interesting isn't it?.. More from Yahoo Answers
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