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In the earlier section we introduced the Wronskian to assist us find out whether two solutions were a fundamental set of solutions. Under this section we will look at the other application of the Wronskian and also an alternate method of computing the Wronskian.
Let's begin with the application. We require introducing a couple of new concepts first.
Specified two non-zero functions f(x) and g(x) write down the subsequent equation
c f ( x ) + k g ( x ) = 0
See that c = 0 and k = 0 will make (1) true for all x regardless of the functions which we use.
Here, if we can get non-zero constants c and k for that (1) will also be true for all x so we call the two functions linearly dependent. Conversely, if the only two constants for that (1) is true are c = 0 and k = 0 so we call the functions linearly independent.
A well-known simple model, applicable for analysing boom-bust cycles in agriculture, but extendable to analysing boom-bust cycles in many different areas of economics is the hog cy
(x^2+x+3)^4
remainder theorem
1. A machine comprises of three transformers A, B and C. Such machine may operate if at least 2 transformers are working. The probability of each transformer working is given as di
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Illustration In a social survey whether the main reason was to establish the intelligence quotient or IQ of resident in a provided area, the given results were acquired as tab
Explain Histogramsin details? Another way to display frequencies is by using a histogram. The following is an example of a histogram using the data from the previous example:
If secA= x+1/4x, prove that secA+tanA=2x or 1/2x. Ans: Sec? = x + 1/4x ⇒ Sec 2 ? =( x + 1/4x) 2 (Sec 2 ?= 1 + Tan 2 ?) Tan 2 ? = ( x +
Finds out the center & radius of each of the following circles & sketch the graph of the circle. a) x 2 + y 2 = 1 b) x 2 + ( y - 3) 2 = 4 Solution In all of these
∫1/sin2x dx = ∫cosec2x dx = 1/2 log[cosec2x - cot2x] + c = 1/2 log[tan x] + c Detailed derivation of ∫cosec x dx = ∫cosec x(cosec x - cot x)/(cosec x - cot x) dx = ∫(cosec 2 x
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