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366_pdfsam_math 54 differential equation solutions odd

# 366_pdfsam_math 54 differential equation solutions odd - to...

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Chapter 6 Since e x is a solution to the associated homogeneous equation and xe x is not, we set s 1 = 1. Since none of the terms x 2 , x , or 1 is a solution to the associated homogeneous equation, we set s 2 = 0. Thus, the form of a particular solution is y p ( x ) = C 1 xe x + C 2 + C 3 x + C 4 x 2 . 3. The associated homogeneous equation for this equation is y + 3 y 4 y = 0. This equation has the corresponding auxiliary equation y 3 + 3 r 2 4 = 0, which, by inspection, has r = 1 as one of its roots. Thus, the auxiliary equation can be factored as follows ( r 1)( r 2 + 4 r + 4) = ( r 1)( r + 2) 2 = 0 . From this we see that the roots to the auxiliary equation are r = 1, 2, 2. Therefore, a general solution to the homogeneous equation is y h ( x ) = c 1 e x + c 2 e 2 x + c 3 xe 2 x . The nonhomogeneous term is g ( x ) = e 2 x . Therefore, a particular solution to the original differential equation has the form y p ( x ) = x s c 1 e 2 x . Since both e 2 x and xe 2 x are solutions
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Unformatted text preview: to the associated homogeneous equation, we set s = 2. (Note that this means that r = − 2 will be a root of multiplicity three of the auxiliary equation associated with the operator equation A [ L [ y ]]( x ) = 0, where A is an annihilator of the nonhomogeneous term g ( x ) = e − 2 x and L is the linear operator L := D 3 + 3 D 2 − 4.) Thus, the form of a particular solution to this equation is y p ( x ) = C 1 x 2 e − 2 x . 5. In the solution to Problem 1, we determined that a general solution to the homogeneous di±erential equation associated with this problem is y h ( x ) = c 1 e x + c 2 e 3 x + c 3 e − 2 x , and that a particular solution has the form y p ( x ) = C 1 xe x + C 2 + C 3 x + C 4 x 2 . 362...
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