97_pdfsam_math 54 differential equation solutions odd

# 97_pdfsam_math 54 differential equation solutions odd - y =...

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Review Problems 11. The right-hand side of this equation is of the form G ( t x )w ith G ( u )=1+cos 2 u .Thu sw e make a substitution t x = u x = t u x 0 =1 u 0 , which yields 1 du dt =1+cos 2 u du dt = cos 2 u sec 2 udu = dt Z sec 2 udu = Z dt tan u = t + C tan( t x )+ t = C. 13. This is a linear equation with P ( x )= 1 /x . Following the method for solving linear equations given on page 51 of the text, we ±nd that an integrating factor µ ( x )=1 /x ,andso d [(1 /x ) y ] dx = 1 x x 2 sin 2 x = x sin 2 x y x = Z x sin 2 xdx = 1 2 x cos 2 x + 1 2 Z cos 2 xdx = 1 2 x cos 2 x + 1 4 sin 2 x + C y = x 2 2 cos 2 x + x 4 sin 2 x + Cx. 15. The right-hand side of the di²erential equation y 0 =2 2 x y + 3 is a function of 2 x
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Unformatted text preview: y = G ( ax + by ) on page 74 of the text. By letting z = 2 x y we can transform the equation into a separable one. To solve, we dierentiate z = 2 x y with respect to x to obtain dz dx = 2 dy dx dy dx = 2 dz dx . Substituting z = 2 x y and y = 2 z into the dierential equation yields 2 dz dx = 2 z + 3 or dz dx = z + 3 . To solve this equation we divide by z + 3, multiply by dx , and integrate to obtain Z ( z + 3) 1 / 2 dz = Z dx 2( z + 3) 1 / 2 = x + C . 93...
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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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