M285_4.1_prelim_linear.pdf - Math 285 4.1 Preliminary Theory in Higher-Order Linear DEs In Chapter 2 we learned to solve some 1st-order DEs that were

M285_4.1_prelim_linear.pdf - Math 285 4.1 Preliminary...

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Math 285: 4.1 Preliminary Theory in Higher-Order Linear DEs In Chapter 2 we learned to solve some 1 st -order DEs that were classified as separable, linear, exact, homogeneous or Bernoulli. Most times (except for the case of 1 st -order linear) we did not find a general solution . Definition: A general solution is a family of solutions defined on an interval I that contains ALL solutions of the DE that are defined on I. GOAL IRU CKaSWeU 4: TR ILQd JeQeUaO VROXWLRQV IRU LLQeaU DE¶V RI RUdeU 2 aQd XS. RECALL: LINEARITY An n th-order linear DE has the form: a n ( x ) y ( n ) ± a n ² 1 ( x ) y ( n ² 1) ± a n ² 2 ( x ) y ( n ² 2) ± ... ± a 2 ( x ) cc y ± a 1 ( x ) c y ± a 0 ( x ) y g ( x ) In addition, we can define an Initial-Value Problem (IVP) for this nth-order linear DE by requiring initial values: y ( x 0 ) y 0 , c y ( x 0 ) y 1 , cc y ( x 0 ) y 2 , . . . , y ( n ² 1) ( x 0 ) y n Theorem 4.1.1: Existence of a Unique Solution for an nth-Order Linear IVP Consider the nth-order linear IVP in the box above. Let the coefficients be continuous on an interval I and let a n ( x ) z 0 for every x in this interval. If x x 0 is any point in this interval, then a solution of the initial-value problem EXISTS on the interval and is UNIQUE. Example 1: a. Given , explain how we know a unique solution exists? b. Say this DE has the general solution: on the interval ²f , f ³ ´ . Solve the IVP. a n ( x ), a n ² 1 ( x ), . . . a 1 ( x ), a 0 ( x ), and g ( x ) y ( x ) cc y ² 3 c y ² 4 y 0 y (0) 1, c y (0) 2 y c 1 e 4 x ± c 2 e ² x continuous coefficient Initial conditions ( Enough ) since the coefficients of y and DH ) its derivatives are continuous and we have 2 initial conditions we can find a unique solution y ( O ) - - c , e. 410 ) t cze - ( D= I at Cz - - I g CH Cz ' I 4C , - Cz = 2 y ' - - 4C , e4X - Cz e - X - 5C 1--3 yl ( O ) : 4C , e. 4107 - Cz e - O = 2 4C I - Cz - - L c , = 31g y=3t C 2=215
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Boundary-Value Problems (BVP) A boundary-value problem is similar to an IVP in that the DE includes prescribed values like: y ( a ) y 0 , y ( b ) y 1 called boundary conditions . It turns out that a BVP my have infinite solutions, a unique solution or no solution. Example 2: Given 2 " 5 ' 8 24 x y xy y ² ± , with family of solutions 2 4 1 2 3 y c x c x ± ± Determine whether a solution can be found that satisfies the following boundary conditions.
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