53_pdfsam_math 54 differential equation solutions odd

53_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.3 s Z 0 d [ µ ( x ) y ]= s Z 0 µ ( x ) xdx µ ( x ) y ( x ) ± ± ± x = s x =0 = s Z 0 µ ( x ) xdx µ ( s ) y ( s ) µ (0) y (0) = s Z 0 µ ( x ) xdx. From the initial condition, y (0) = 2. Also, note that µ (0) = exp 0 Z 0 p 1+sin 2 tdt = e 0 =1 . This yields µ (0) y (0) = 2 and so µ ( s ) y ( s )= s Z 0 µ ( x ) xdx +2 . Dividing by µ ( s ) and interchanging x and s give the required. (b) The values of µ ( x ), x =0 . 1, 0 . 2, ... ,1 . 0, approximated by using Simpson’s rule, are given in Table 2-C. Table 2–C : Approximations of ν ( x )= R x 0 1+sin 2 tdt and µ ( x )= e ν ( x ) using Simpson’s rule. x ν ( x ) ν ( x ) ν x ) µ ( x ) µ ( x ) µ ( x ) x ν ( x ) ν ( x ) ( x ) µ ( x ) µ ( x ) µ ( x ) 0.0 0.0 1.0000 0.6
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