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Consider the subsequent IVP.
y' = f(t,y) , y(t0) = y0
If f(t,y) and ∂f/∂y are continuous functions in several rectangle a < t < b, g < y < d, containing the point (to, yo) then there is a unique solution to the IVP in some interval to - h < t < to + h which is included in a < t < b.
That's it. Unlike the first theorem, this one cannot really be used to find an interval of validity. Thus, we will know that a unique solution exists if the conditions of the theorem are met, but we will in fact need the solution in order to find out its interval of validity. Remember as well that for non-linear differential equations it emerges that the value of y0 may influence the interval of validity.
Here is an illustration of the problems that can happen when the conditions of this theorem are not met.
0.875 of a number is 2282. What is the number ?
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how to slove problems on equations
Domain of a Vector Function There is a Vector function of a single variable in R 2 and R 3 have the form, r → (t) = {f (t), g(t)} r → (t) = {f (t) , g(t), h(t)} co
Solve for x , y (x + y - 8)/2 =( x + 2 y - 14)/3 = (3 x + y - 12 )/ 11 (Ans: x=2, y=6) Ans : x+ y - 8/2 = x + 2y - 14 /3 = 3x+ y- 12/11
There is one final topic that we need to address as far as solution sets go before leaving this section. Consider the following equation and inequality.
Solve : 4x2+2x+3=0 Ans) x^2 + (1/2)x = -(3/4) (x+1/4)^2 = 1/16 - 3/4 = -11/16 implies x = (-1+i(11)^(1/2))/4 and its conjugate.
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Now we have to look at rational expressions. A rational expression is a fraction wherein the numerator and/or the denominator are polynomials. Here are some examples of rational e
5:9 and 3:5 then find a:b:c?
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