HW4_Soln_TL

# HW4_Soln_TL - Solution of HW4 3.1.4 3.1.5 Same as above the...

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Solution of HW4 3.1.4 3.1.5 Same as above, the characteristic equation would be 2 5 0 r r . The roots are r=0, -5. Hence the general solution is ( 5 ) 1 2 t c y c e . 3.1.16 3.1.17 Since the general solution is 2 3 1 2 t t e y c c e , the characteristic equation would therefore has the form 2 (r 2)(r 3) 6 0 r r , the corresponding differential equation is '' ' 6y 0 y y .

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3.1.21 The characteristic equation is 2 2 (r 2)(r 1) 0 r r , so the solution is 2, -1 r . Then the general solution is 2 1 2 t t c y e c e , substitute the initial value condition into the solution we have, 1 2 1 2 (0) c '(0) 2c c 2 y c y Hence, 1 2 3 c , 2 2 2 3 c . So 2 2 2 3 3 2 t t y e e . Since when t   , 0 y , 2   . 3.2.2 2 2 cos sin (cost,sint) sin sin cos cos 1 t t W t t t t 3.2.9 3.2.22 According to theorem 3.2.5, the fundamental set of solutions are solutions of IVP shown as below: 1 1 1 1 1 '' ' 2 0 (0) 1 '(0) 0 y y y y y 2 2 2 2 2 '' ' 2 0 (0) 0 '(0) 1 y y y y y

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