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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.
1. What is the present value of a security that will pay $15,000 in 15 years if securities of equal risk pay 8.9% annually? Round your answer to the nearest cent. 475,858.20
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The general solution of the differential equation (dy/dx) +x^2 = x^2*e^(3y). Solution)(dy/dx) +x^2 = x^2*e^(3y) dy/dx=x 2 (e 3y -1) x 2 dx=dy/(e 3y -1) this is an elementar
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solve by factorization method; 10x-6y-3z=100, -6x+10y-5z=100, -3x-5y+10z=100
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Grouped Data For grouped data of a paired population where, f is the
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2sqrt73x
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