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89_pdfsam_math 54 differential equation solutions odd

# 89_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.6 µ ( t ) = exp 2 t dt = t 2 d ( t 2 u ) dt = 2 t t 2 u = 2 t dt = 2 ln | t | + C u = 2 t 2 ln | t | + Ct 2 x 2 = 2 t 2 ln | t | + Ct 2 . x 0 is also a solution, which we lost dividing the equation by x 3 . 27. This equation is a Bernoulli equation with n = 2, because it can be written in the form dr 2 θ r = r 2 θ 2 . Dividing by r 2 and making the substitution u = r 1 , we obtain a linear equation. r 2 dr 2 θ r 1 = θ 2 du 2 θ u = θ 2 du + 2 θ u = θ 2 µ ( θ ) = exp 2 θ = θ 2 d ( θ 2 u ) = 1 θ 2 u = θ + C u = θ + C θ 2 . Making back substitution (and adding the lost solution r 0), we obtain a general solution r = θ 2 C
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