219_pdfsam_math 54 differential equation solutions odd

219_pdfsam_math 54 differential equation solutions odd - y...

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Exercises 4.6 Thus, y 0 1 ( t )= e t and y 0 2 ( t )= te t + e t . This means that we want to solve the system (see system (9) on page 195 of text) e t v 0 1 + te t v 0 2 =0 , e t v 0 1 + ( te t + e t ) v 0 2 = t 1 e t . Subtracting these two equations yields e t v 0 2 = t 1 e t v 0 2 = t 1 . So v 2 ( t )= Z t 1 dt =ln | t | + c 3 . Also, we have from the Frst equation of the system e t v 0 1 = te t v 0 2 = te t t 1 = e t v 0 1 = 1 . So, v 1 ( t )= t + c 4 . By letting c 3 and c 4 equal to zero, and plugging the expressions found above for v 1 ( t )and v 2 ( t ) into the equation deFning y p ( t ) , we obtain a particular solution y p ( t )= te t + te t ln | t | . We obtain a general solution of the nonhomogeneous equation by adding this expression for
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Unformatted text preview: y p ( t ) to the expression for y h ( t ). Thus, we obtain y ( t ) = c 1 e t + c 2 te t te t + te t ln | t | = c 1 e t + ( c 2 1) te t + te t ln | t | . If we let C 1 = c 1 and C 2 = c 2 1, we can express this general solution in the form y ( t ) = C 1 e t + C 2 te t + te t ln | t | . 215...
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