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The actual solution is the specific solution to a differential equation which not only satisfies the differential equation, although also satisfies the specified initial conditions.
Illustration: What is the actual solution to the subsequent IVP?
2ty' + 4y = 3; y(1) = -4
Solution: This is in fact easier to do than this might at first appear. From the earlier illustration we already identify that differential equations have all solutions are of the form:
y(t) = 3/4 + c/t2
All that we require to do is find out the value of c that will provide us the solution that we're after. To determine this all we require do is utilize our initial condition that are given as:
-4 = y(1) = 3/4 + c/12
c= -4 -3/4 = -19/4
Thus, the actual solution to the Initial Value Problem is:
y(t) = ¾ - 19/4t2
From this last illustration we can notice that once we have the general solution to a differential equation determining the actual solution is nothing more than applying the initial conditions and resolving for the constants which are in the general solution.
1+2x
if 1/x+2, 1/x+3, 1/x+5 are in AP find x Ans 1/x+2,1/x+3, 1/x+5 are in AP find x. 1/x+3 - 1/x+2 = 1/x+5-1/x+3 => 1/x 2 +5x+6 = 2/ x 2 +8x +15 => On solving we get x
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If α & ß are the zeroes of the polynomial 2x 2 - 4x + 5, then find the value of a.α 2 + ß 2 b. 1/ α + 1/ ß c. (α - ß) 2 d. 1/α 2 + 1/ß 2 e. α 3 + ß 3 (Ans:-1, 4/5 ,-6,
y''''+6y''+10y=10x^(4)+24x^(3)+2x^(2)-12x+18
scope of operation research and its limitations
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