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In this theorem we identify that for a specified differential equation a set of fundamental solutions will exist.
Consider the differential equation
y′′ + p (t ) y′ + q (t ) y = 0
Here p(t) and q(t) are continuous functions on any interval I. select t0 to be any point in the interval I. Let y1(t) be a solution to the differential equation which satisfies the initial conditions.
y(t0) = 1
y′ (t0) = 0
Let y2(t) be a solution to the differential equation which satisfies the initial conditions.
y (t0) = 0
y′ (t0) = 1
So y1(t) and y2(t) form a fundamental set of solutions for the differential equation.
This is easy enough to illustrate that these two solutions form a fundamental set of solutions. Just calculate the Wronskian.
Thus, fundamental sets of solutions will exist; we can solve the two IVP's specified in the theorem.
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Examples on Log rules: Example: Calculate (1/3)log 10 2. Solution: log b n√A = log b A 1/n = (1/n)log b A (1/3)log 10 2 = log 10 3 √2 = log 10 1.
Solve the subsequent differential equation and find out the interval of validity for the solution. Let's start things off along with a fairly simple illustration so we can notic
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To solve out linear equations we will make heavy use of the following facts. 1. If a = b then a + c = b + c for any c. All it is saying that we can add number, c, to both sides
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A right triangle whose sides are 15 cm and 20 cm is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Ans : 3768cu.cm,1318.8
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