206_pdfsam_math 54 differential equation solutions odd

# 206_pdfsam_math 54 differential equation solutions odd - c...

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Chapter 4 (ii) For the equation y 0 y 0 2 y =s in2 x, a particular solution has the form y p, 2 ( x )= A cos 2 x + B sin 2 x . Substitution yields ( 4 A cos 2 x 4 B sin 2 x ) ( 2 A sin 2 x +2 B cos 2 x ) 2( A cos 2 x + B sin 2 x ) =( 6 A 2 B )cos2 x +(2 A 6 B )sin2 x =s in2 x 6 A 2 B =0 , 2 A 6 B =1 A =1 / 20 , B = 3 / 20 . So, y p, 2 ( x )= 1 20 cos 2 x 3 20 sin 2 x. Therefore, a general solution to the original equation is y ( x )= y p, 1 ( x ) y p, 2 ( x )+ y h ( x ) = 3 10 cos x 1 10 sin x 1 20 cos 2 x + 3 20 sin 2 x + c 1 e x + c 2 e 2 x . Next, we ±nd c 1 and c 2 such that the initial conditions are satis±ed. 7 / 20 = y (0) = 3 / 10 1 / 20 + c 1 + c 2 , 1 / 5= y 0 (0) = 1 / 10 + 2(3 / 20) c 1 +2 c 2 c 1 + c 2 =0 ,
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Unformatted text preview: c 1 + 2 c 2 = 0 ⇒ c 1 = 0 , c 2 = 0 . With these constants, the solution becomes y ( x ) = − 3 10 cos x − 1 10 sin x − 1 20 cos 2 x + 3 20 sin 2 x . 29. The roots of the auxiliary equation, r 2 − 1 = 0, are r = ± 1. Therefore, a general solution to the corresponding homogeneous equation is y h ( θ ) = c 1 e θ + c 2 e − θ . (i) For the equation y − y = sin θ, 202...
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