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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.
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Smith keeps track of poor work. Often on afternoon it is 5%. If he checks 300 of 7500 instruments what is probability he will find less than 20 substandard?
Test Of Hypothesis On Proportions It follows a similar method to the one for means except that the standard error utilized in this case: Sp = √(pq/n) Z score is computed
Previous year's budget was 12.5 million dollars. This year's budget is 14.1 million dollars. How much did the budget increase? Last year's budget must be subtracted from this y
What are Natural Numbers and Whole Numbers? Natural numbers are the numbers that you "naturally" use for counting: 1, 2, 3, 4, ... The set of whole numbers is the set of
Stratified sampling In stratified sampling case the population is divided into groups in such a way that units in each group are as same as possible in a process called strati
Find out the roots of the following quadratic equation. 3x 2 + 7x = 0 Solution: Using Equation 6, one root is determined. x = 0 Using Equation 7, substitute the
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Survey 83% of community for a park. Randomly select 21 people if they do or do not want a park. Can you use normal distribution to approximate binomial distribution?If so find mean
from 0->1: Int sqrt(1-x^2) Solution) I=∫sqrt(1-x 2 )dx = sqrt(1-x 2 )∫dx - ∫{(-2x)/2sqrt(1-x 2 )}∫dx ---->(INTEGRATION BY PARTS) = x√(1-x 2 ) - ∫-x 2 /√(1-x 2 ) Let
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