chapter2.page04

# chapter2.page04 - C = 3 Hence x t = t 2 3 Notice if we drop...

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4 1. EXPLICITLY SOLVABLE FIRST ORDER DIFFERENTIAL EQUATIONS any region outside the bounding box , if not in the auto execution mode hit [F9] too, to ﬁnd the antiderivative Z x + 2 - e 3 x - sin (3 x ) dx 2 3 ( x + 2) 3 2 - 1 3 exp 3 x + 1 3 cos(3 x ) . Here exp represent exponential function. From Calculus, we know that Z x + 2 - e 3 x - sin (3 x ) dx = 2 3 ( x + 2) 3 2 - 1 3 exp 3 x + 1 3 cos(3 x ) + C. Mathcad omits the C for believing you know it! Remember C is im- portant when we try to ﬁnd a particular solution for a given, so called, ”initial” condition. Example 1.3 . Find a function x ( t ) such that x 0 ( t ) = 2 t and x (0) = 3 . Solution If we take antiderivative on both sides, we have Z x 0 ( t ) dt = Z 2 tdt which leads to x ( t ) = t 2 + C. (1) Evaluate this x ( t ) at t = 0 we have x (0) = 0 2 + C = C So the requirement x (0) = 3 implies
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Unformatted text preview: C = 3. Hence x ( t ) = t 2 + 3 . Notice if we drop the C from (1), we would not be able to ﬁnd x ( t ) that satisﬁes x (0) = 3! a The problem of ﬁnding a particular solution with initial condition is called initial value problem, which we will discuss in the later section. a Practice 1. Find the following deﬁnite integral using Mathcad . (a) R 2-1 x 3-x cos( x 2 + 1) dx (b) R 4 1 x 3 +4 x-5 x +1 dx (c) R-5 e 3 x-x x 2 +3 dx (d) R 10-1 x 2-x sec( x 2 + 1) dx 2. Find the following indeﬁnite integral using Mathcad . (a) R x 5-x p ( x 2 + 1) dx (b) R x 3 +4 x-5 x 2 +1 dx...
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