chapter2.page14

chapter2.page14 - a solution For the linear first order...

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14 1. EXPLICITLY SOLVABLE FIRST ORDER DIFFERENTIAL EQUATIONS Here is how to get the graph in Mathcad : - we first defined a function x ( t,C ):= 0 . 2 0 . 02+ Ce - 0 . 2 t with two argu- ments t and C by typing x(t,C):0.2/0.02+C*e^-0.2t , - then we defined a range variable t:=0,0.1 ... 20 by typing t:0,0.1;20 , - finally, type @ at a blank area to get the xy-plot; in the function place holder, we put x(t,0),x(t,-0.01),x(t,-0.008),x(t,0.004),x(t,0.01) and type t in the variable place holder, and click at outside of the box to get the graph. a 4. Initial value problem and existence theorem The problem of finding a particular solution x ( t ) of an ODE x 0 = f ( t,x ) that satisfying condition x ( t 0 ) = x 0 is called an initial value problem. The problem can be written x 0 = f ( t,x ) x ( t 0 ) = x 0 Since solving an ODE requires a lots of hard work, it is wise to know if a given equation has any solution before march on the journey of find
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Unformatted text preview: a solution. For the linear first order ODE x + a ( t ) x = b ( t ) we following existence-uniqueness theorem. Theorem 4.1 . If the function a ( t ) and b ( t ) are continuous on the open interval I containing the point t , then the initial value problem x + a ( t ) x = b ( t ) , x ( t ) = x has a unique solution x ( t ) on I, given by x ( t ) = e R a ( t ) dt ˆ Z b ( t ) e-R a ( t ) dt dt + C ! (8) with an appropriate value of C. So solving an initial value problem for linear first order ODE is easy. One just apply the formula (8) and using the equation x ( t ) = x to find C . Example 4.1 . Solve the initial value problem x +2 tx = 4 t 3 , x (0) = 1 . Solution First we apply the formula (8) x ( t ) = e R a ( t ) dt ˆ Z b ( t ) e-R a ( t ) dt dt + C ! ,...
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This note was uploaded on 10/04/2011 for the course MAP 4231 taught by Professor Dr.han during the Spring '11 term at UNF.

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