HW4 Solutions(1).pdf - ESE 351-01 Fall 2017 Homework Set#4 due Sep 26 When finding solutions y(t solve only for t > 0 4.1 Mukai 4.9.1(b 4.2 Mukai

HW4 Solutions(1).pdf - ESE 351-01 Fall 2017 Homework Set#4...

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ESE 351-01, Fall 2017 Homework Set #4, due Sep. 26 When finding solutions, y ( t ), solve only for t > 0. 4.1 Mukai 4.9.1(b) 4.2 Mukai 4.9.2(c) 4.3 Mukai 4.9.5(b) 4.4 Consider this 1 st order system, where T and E are constants. T d dt y + y = Eu u t ( ) = δ t ( ) y 0 ( ) = y 0 a) Indicate (by corresponding canonical function) the class of each of the following 3 terms. D y y D 1 y b) What is y (0 + )? c) What is y ( t ) for t > 0? d) Use results from above to solve for the head, y , in this fluid system when u ( t ) = δ ( t ) and y (0 - ) = 0. Solve using the methods of Mukai Chapter 4.
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ESE 351-01, Fall 2017 Homework Set #4, due Sep. 26 Solutions 4.1 Mukai 4.9.1(b) D 2 y + 5 D y + 6 y = D 2 u + 10 D u + 25 u y 0 ( ) = 0 D y 0 ( ) = 0 u t ( ) = 1 t ( ) Find class of the leading term and then integrate to find other terms as needed. D 2 y + 5 D y + 6 y = δ 1 ( ) t ( ) + 10 δ t ( ) + 25 1 t ( ) D 2 y t ( ) δ 1 ( ) t ( ) D y t ( ) δ t ( ) y t ( ) 1 t ( ) D 1 y t ( ) r t ( ) Integrate the original equation from 0 - to 0 + . D 2 y + 5 D y + 6 y = δ 1 ( ) t ( ) + 10 δ t ( ) + 25 1 t ( ) D 2 ydt 0 0 + + 5 D ydt 0 0 + + 6 ydt 0 0 + 0 ! = δ 1 ( ) t ( ) dt 0 0 + 0 ! " # $ # + 10 δ t ( ) dt 0 0 + 1 ! " # $ # + 25 1 t ( ) dt 0 0 + 0 !" # $ # D y 0 + ( ) D y 0 ( ) + 5 y 0 + ( ) y 0 ( ) = 10 Δ D y + 5 Δ y = 10 The integrands of the 2 left-most terms are singular, so there is no guarantee that their definite integrals are 0. We have 2 values to solve for, and only 1 equation so far. Integrate (indefinitely) the original equation and again take the definite integral of the result from 0 - to 0 + . D y + 5 y + 6 D 1 y = δ t ( ) + 10 1 t ( ) + 25 r t ( ) D ydt 0 0 + + 5 ydt 0 0 + 0 ! + 6 D 1 ydt 0 0 + 0 ! " # $ # = δ dt 0 0 + 1 !
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