92_pdfsam_math 54 differential equation solutions odd

92_pdfsam_math 54 differential equation solutions odd -...

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Chapter 2 Z 3 zdz 3 z 2 +1 Z dz 3 z 2 +1 = ln | v | + C 1 1 2 ln ( 3 z 2 +1 ) 1 3 arctan ± z 3 ² = ln | v | + C 1 ln ³ (3 z 2 +1) v 2 ´ 2 3 arctan ± z 3 ² = C 2 . Making back substitution, after some algebra we get ln ³ 3( t 1) 2 +( x +3) 2 ´ + 2 3 arctan µ x +3 3( t 1) = C. 37. In Problem 5, we have written the equation in the form dy dx y = e 2 x y 3 y 3 dy dx y 2 = e 2 x . Making a substitution u = y 2 (and so u 0 = 2 y 3 y 0 ) in this Bernoulli equation, we get du dx +2 u = 2 e 2 x µ ( x )=exp ·Z 2 dx ¸ = e 2 x d ( e 2 x u ) dx = 2 e 2 x e 2 x = 2 e 4 x e 2 x u = Z ( 2 e 4 x ) dx = 1 2 e 4 x + C u = 1 2 e 2 x + Ce 2 x y 2 = 1 2 e 2 x + Ce 2 x . The constant function y 0 is also a solution, which we lost dividing the equation by y 3 . 39. Since the equation is homogeneous, we make a substitution
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