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207_pdfsam_math 54 differential equation solutions odd

# 207_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.5 a particular solution has the form y p, 1 ( x ) = A cos θ + B sin θ . Substitution into the equation yields ( A cos θ B sin θ ) ( A cos θ + B sin θ ) = 2 A cos θ 2 B sin θ = sin θ 2 A = 0 , 2 B = 1 A = 0 , B = 1 / 2 . So, y p, 1 ( θ ) = (1 / 2) sin θ . (ii) For the equation y y = e 2 θ , a particular solution has the form y p, 2 ( θ ) = Ae 2 θ . Substitution yields ( Ae 2 θ ) ( Ae 2 θ ) = 3 Ae 2 θ = e 2 θ A = 1 / 3 , and y p, 2 ( θ ) = (1 / 3) e 2 θ . By the superposition principle, a particular solution to the original nonhomogeneous equation is given by y p ( θ ) = y p, 1 ( θ ) y p, 2 ( θ ) = (1 / 2) sin θ (1 / 3) e 2 θ , and a general solution is y ( θ ) = y p ( θ ) + y
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