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229_pdfsam_math 54 differential equation solutions odd

# 229_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.6 Thus y p ( t ) = ( t + 2) e t e t t ( t + 1) = t 2 2 t 2 . (Note that 2 t 2 = 2( t + 1) = 2 y 2 ( t ) is a solution to the corresponding homogeneous equation. Thus, t 2 = y p ( t ) + 2 y 2 ( t ) is another particular solution.) 23. We are seeking for a particular solution to the given equation of the form y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) = v 1 ( t )(5 t 1) + v 2 ( t ) e 5 t . Since y 1 ( t ) 5, y 2 ( t ) = 5 e 5 t , the system (9), with a = a ( t ) = t and g ( t ) = t 2 e 5 t , becomes v 1 ( t )(5 t 1) + v 2 ( t ) e 5 t = 0 , 5 v 1 ( t ) 5 v 2 ( t ) e 5 t = t 2 e 5 t t = te 5 t . Dividing the second equation by 5 and adding to the first equation yields 5 tv 1 ( t ) = 1 5 te 5 t v 1 ( t ) = 1 25 e 5 t v 1 ( t ) = 1 125 e 5 t . Substituting v 1 ( t ) into the first equation, we get 1 25 e 5 t (5 t 1) + v 2 ( t ) e 5 t = 0 v 2 ( t ) = 5 t 1 25 v 2 ( t ) = t 2 10 + t 25 . Thus y p ( t ) = 1 125 e 5 t (5 t 1) + t 2 10 + t 25 e 5 t = 1 125 t 2 10 e 5 t . (Since (1 / 125) e 5 t = (1 / 125) y 2 ( t ) is a solution to the corresponding homogeneous equation,
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Unformatted text preview: the function − ( t 2 / 10) e − 5 t is also a particular solution.) 25. A general solution to the corresponding homogeneous equation is y h ( x ) = c 1 y 1 ( x ) + c 2 y 2 ( x ) = x − 1 / 2 ( c 1 cos x + c 2 sin x ) . To Fnd a particular solution to the original equation, we apply the method of variation of parameters. To form the system (9) on page 195, we need y 1 and y 2 . Applying the product rule, we get y 1 ( x ) = − 1 2 x − 3 / 2 cos x − x − 1 / 2 sin x, 225...
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