350_pdfsam_math 54 differential equation solutions odd

350_pdfsam_math 54 differential equation solutions odd -...

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Chapter 6 To satisfy the initial conditions, first we find y ( x ) = 1 x + C 1 + C 2 (ln x + 1) + C 3 (ln x ) 2 + 2 ln x , y ( x ) = 1 x 2 + C 2 x + C 3 2 ln x x + 2 x . Substituting the initial conditions, y (1) = 3, y (1) = 3, and y (1) = 0, we get the system 3 = y (1) = C 1 , 3 = y (1) = 1 + C 1 + C 2 , 0 = y (1) = 1 + C 2 + 2 C 3 C 1 = 3 , C 1 + C 2 = 2 , C 2 + 2 C 3 = 1 C 1 = 3 , C 2 = 1 , C 3 = 1 . Thus, y ( x ) = ln x + 3 x x ln x + x (ln x ) 2 is the desired solution. 23. Substituting y 1 ( x ) = sin x and y 2 ( x ) = x into the given differential operator yields L [sin x ] = (sin x ) + ( sinx ) + x (sin x ) = cos x + cos x + x sin x = x sin x, L [ x ] = ( x ) + ( x ) + x ( x ) = 0 + 1 + x 2 = x 2 + 1 . Note that L [ y ] is a linear operator of the form (7). So, we can use the superposition principle.
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