m3354_f05_q4_sol

m3354_f05_q4_sol - dy y = dx x ln( y ) = ln( x ) + c Thus y...

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Math 3354 Sec. 2, Quiz # 4 Name KEY 1. ( x missing ) Find the general solution of x 00 = x 0 . y = x 0 y 0 = y so dy y = dt ln( y ) = t + c 1 Thus y = c 2 e t and x 0 = c 2 e t so that x = c 2 e t + c 3 2. ( t missing [chain rule] ) Find the general solution of xx 00 = ( x 0 ) 2 . y = x 0 xy dy dx = y 2
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Unformatted text preview: dy y = dx x ln( y ) = ln( x ) + c Thus y = C 1 x x = C 1 x so dx x = C 1 dt ln( x ) = C 1 t + C 2 Finally x = C 2 e C 1 t...
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