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383_pdfsam_math 54 differential equation solutions odd

383_pdfsam_math 54 differential equation solutions odd -...

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Exercises 6.4 Therefore, we have y p ( x ) = v 1 ( x ) + v 2 ( x ) cos x + v 3 ( x ) sin x = ln(sec x ) + cos 2 x + sin 2 x sin x ln(sec x + tan x ) y p ( x ) = ln(sec x ) sin x ln(sec x + tan x ) + 1 . Since y 1 is a solution to the homogeneous equation, we may choose y p ( x ) = ln(sec x ) sin x ln(sec x + tan x ) . Note: We left the absolute value signs off ln(sec x ) and ln(sec x + tan x ) because of the stated domain: 0 < x < π/ 2. 7. First, we divide the differential equation by x 3 to obtain the standard form y 3 x 1 y + 6 x 2 y 6 x 3 y = x 4 , x > 0 , from which we see that g ( x ) = x 4 . Given that { x, x 2 , x 3 } is a fundamental solution set for the corresponding homogeneous equation, we are looking for a particular solution of the form
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