369_pdfsam_math 54 differential equation solutions odd

# 369_pdfsam_math 54 differential equation solutions odd - we...

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Exercises 6.3 By substituting these expressions into the original equation, we obtain y 0 p 3 y 0 p +3 y 0 p y = e x ± A ( x 3 +9 x 2 +18 x +6 ) e x ² 3 ± A ( x 3 +6 x 2 +6 x ) e x ² +3 ± A ( x 3 +3 x 2 ) e x ² Ax 3 e x = e x 6 Ae x = e x A = 1 6 , and so y p ( x )= x 3 e x / 6. A general solution to the given equation then has the form y ( x )= y h ( x )+ y p ( x )= c 1 e x + c 2 xe x + c 3 x 2 e x + 1 6 x 3 e x . 11. The operator D 5 , that is, the Ffth derivative operator, annihilates any polynomial of degree at most four. In particular, D 5 annihilates the polynomial x 4 x 2 + 11. 13. According to (i) on page 334 of the text, the operator [ D ( 7)] = ( D + 7) annihilates the exponential function e 7 x . 15. The operator ( D 2) annihilates the function f 1 ( x ):= e 2 x and the operator ( D 1) annihilates the function f 2 ( x ):= e x . Thus, the composition of these operators, namely, ( D 2)( D 1), annihilates both of these functions and so, by linearity, it annihilates their algebraic sum. 17. This function has the same form as the functions given in (iv) on page 334 of the text. Here
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Unformatted text preview: we see that α = − 1, β = 2, and m − 1 = 2. Thus, the operator ± ( D − {− 1 } ) 2 + 2 2 ² 3 = ± ( D + 1) 2 + 4 ² 3 annihilates this function. 19. Given function as a sum of two functions. The Frst term, xe − 2 x , is of the type (ii) on the page 334 of the text with m = 2 and r = − 2; so [ D − ( − 2)] 2 = ( D + 2) 2 annihilates this function. The second term, xe − 5 x sin 3 x , is annihilated by ± ( D − ( − 5)) 2 + 3 2 ² 2 = ± ( D + 5) 2 + 9 ² 2 according to (iv). Therefore, the composition [( D + 2) 2 ( D + 5) 2 + 9] 2 annihilates the function xe − 2 x + xe − 5 x sin 3 x . 365...
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