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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 ) with G ( u ) = 1 + cos 2 u . Thus we make a substitution t x = u x = t u x = 1 u , which yields 1 du dt = 1 + cos 2 u du dt = cos 2 u sec 2 u du = dt sec 2 u du = 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 find that an integrating factor µ ( x ) = 1 /x , and so d [(1 /x ) y ] dx = 1 x x 2 sin 2 x = x sin 2 x y x = x sin 2 x dx = 1 2 x cos 2 x + 1 2 cos 2 x dx = 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 differential equation y = 2 2 x y + 3 is a function of 2 x y and so can be solved using the method for equations of the form
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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 di²erentiate 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 di²erential 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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