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 =s in 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 0 y = e 2 θ , a particular solution has the form y p, 2 ( θ )= Ae 2 θ . Substitution yields ( Ae 2 θ ) 0 ( 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 h (
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