201_pdfsam_math 54 differential equation solutions odd

# 201_pdfsam_math 54 differential equation solutions odd - y...

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Exercises 4.5 3. The corresponding homogeneous equation, y 0 y = 0, has the associated auxiliary equation r 2 1=( r 1)( r + 1) = 0. This gives r = ± 1 as the roots of this equation, and a general solution to the homogeneous equation is y h ( t )= c 1 e t + c 2 e t . Combining this solution with the particular solution, y p ( t )= t , we Fnd that a general solution is given by y ( t )= y p ( t )+ y h ( t )= t + c 1 e t + c 2 e t . 5. The corresponding auxiliary equation, r 2 r 2 = 0, has the roots r = 1, 2. Hence, a general solution to the corresponding homogeneous equation is θ h ( t )= c 1 e 2 t + c 2 e t .Byth e superposition principle, a general solution to the original nonhomogeneous equation is θ ( t )= θ p ( t )+ θ h ( t )= t 1+ c 1 e 2 t + c 2 e t . 7. ±irst, we rewrite the equation in standard form, that is, y 0 2 y 0 + y =2 e x . The corresponding homogeneous equation, y 0 2 y 0 + y = 0, has the associated auxiliary equation r 2 2 r +1=( r 1) 2 =0 . Thus r
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Unformatted text preview: y h ( x ) = c 1 xe x + c 2 e x . Combining this with the particular solution, y p ( x ) = x 2 e x , we Fnd that a general solution is given by y ( x ) = y p ( x ) + y h ( x ) = x 2 e x + c 1 xe x + c 2 e x . 9. We can write the nonhomogeneous term as a dierence t 2 + 4 t t 2 e t sin t = ( t 2 + 4 t ) ( t 2 e t sin t ) = g 1 ( t ) g 2 ( t ) . Both, g 1 ( t ) and g 2 ( t ), have a form suitable for the method of undetermined coecients. There-fore, we can apply this method to Fnd particular solutions y p, 1 ( t ) and y p, 2 ( t ) to 3 y + 2 y + 8 y = g 1 ( t ) a n d 3 y + 2 y + 8 y = g 2 ( t ) , respectively. Then, by the superposition principle, y p ( t ) = y p, 1 ( t ) y p, 2 ( t ) is a particular solution to the given equation. 197...
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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