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# 4.3we - 4.3 THE METHOD OF UNDETERMINED COEFFICIENTS KIAM...

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4.3: THE METHOD OF UNDETERMINED COEFFICIENTS KIAM HEONG KWA Recall that we applied the method of undetermined coefficients to solve second-order linear equations in section 3.6. The same method can also be applied to solve the higher order linear equation (1) a 0 y ( n ) + a 1 y ( n - 1) + · · · + a n y = g ( t ) , where a i R , i = 0 , 1 , · · · , n , with a 0 6 = 0, and g ( t ) is a finite sum of functions, each of which being the product of polynomials, an expo- nential, and a cosine or a sine function. To be explicit, say g ( t ) = g 1 ( t ) + g 2 ( t ) + · · · + g m ( t ) . For each j , j = 1 , 2 , · · · , m , let P j ( t ) denote a polynomial of degree n j and let α, β R . Let (a) Y j ( t ) = t s j ( A 0 j t n j + A 1 j t n j - 1 + · · · + A n j j ) , where A ij R , i = 0 , 1 , · · · , n j , provided g j ( t ) = P j ( t ); (b) Y j ( t ) = t s j e αt ( A 0 j t n j + A 1 j t n j - 1 + · · · + A n j j ) , where A ij R , i = 0 , 1 , · · · , n j , provided g j ( t ) = P j ( t ) e αt ; and (c) Y j ( t ) = t s j e αt [( A 0 j t n j + A 1 j t n j - 1 + · · · + A n j j ) cos βt + ( B 0 j t n j + B 1 j t n j - 1 + · · · + B n j j ) sin βt ] , where A ij , B ij R , i = 0 , 1 , · · · , n j , provided either g j ( t ) = P j ( t ) e αt cos βt or g j ( t ) = P j ( t ) e αt sin βt . In each case, s j denotes the least nonnegative integer such that no term in each Y j ( t ), j = 1 , 2 , · · · , m , is dependent on any solution of the homogeneous equation corresponding to (1). For an n th order Date : January 30, 2011. 1

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2 KIAM HEONG KWA equations, 0 s j n for each j , j = 1 , 2 , · · · , m . A particular solution of (1) is then given by Y ( t ) = Y 1 ( t ) + Y 2 ( t ) + · · · + Y m ( t ) . Finally, the constants A ij ’s and B ij ’s can be found by inserting Y ( t ) into (1) and comparing the cofficients of the like terms in the resulting equation. Worked Examples. Example 1 (Problem 11 in the text) . The solution of the initial value problem (2) y 000 - 3 y 00 + 2 y 0 = t + e t with (3) y (0) = 1 , y 0 (0) = - 1 4 , and y 00 (0) = - 3 2 is (4) y ( t ) = 1 + 1 4 t 2 + 3 4 t - te t . Solution. Consider the homogeneous equation that corresponds to (2): y 000 - 3 y 00 + 2 y 0 = 0 .
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4.3we - 4.3 THE METHOD OF UNDETERMINED COEFFICIENTS KIAM...

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