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105_pdfsam_math 54 differential equation solutions odd

105_pdfsam_math 54 differential equation solutions odd -...

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Review Problems 1 2 ln ( z 2 + 2 ) + 1 2 arctan z 2 = ln | u | + C 1 ln ( z 2 + 2 ) u 2 + 2 arctan z 2 = C ln ( v 2 + 2 u 2 ) + 2 arctan v u 2 = C ln ( y 2) 2 + 2( x 1) 2 + 2 arctan y 2 ( x 1) 2 = C. The initial condition, y (0) = 2, gives C = ln 2, and so the answer is ln ( y 2) 2 + 2( x 1) 2 + 2 arctan y 2 ( x 1) 2 = ln 2 . 39. Multiplying the equation by y , we get y dy dx 2 x y 2 = 1 x . We substitute u = y 2 and obtain 1 2 du dx 2 x u = 1 x du dx 4 x u = 2 x , which is linear and has an integrating factor
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