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Theorem
Consider the subsequent IVP.
y′ = p (t ) y = g (t )
y (t0)= y0
If p(t) and g(t) are continuous functions upon an open interval a < t < b and the interval includes to, after that there is a unique solution to the IVP on such interval.
Therefore, just what does this theorem tell us? Initially, it tells us that for nice adequate linear first order differential equations solutions are guaranteed to exist and more significantly the solution will be particular. We may not be capable to get the solution, but do identify that it exists and which there will only be one of them. It is the very significant aspect of this theorem. Identifying that a differential equation has a unique solution is probably more significant than actually having the solution itself!
Subsequently, if the interval in the theorem is the largest possible interval on that p(t) and g(t) are continuous so the interval is the interval of validity for the solution. This means that for linear first order differential equations, we won't want to actually solve the differential equation in order to get the interval of validity. See that the interval of validity will based only partially on the initial condition. The interval should hold to, but the value of yo, has no consequence on the interval of validity.
Determine or find out if the sets of vectors are parallel or not. (a) a → = (2,-4,1), b = (-6, 12 , -3) (b) a → = (4,10), b = (2,9) Solution (a) These two vectors
LAST COST METHOD
is 1 and 1/2+2 and 1/7 3 and 9/4
A drug has a decay rate of k = - ¼ ln(¾) / hr. How soon after an initial dose of 1600 mg will the drug reach its minimum therapeutic value of 900 mg in the body?
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