205_pdfsam_math 54 differential equation solutions odd

# 205_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.5 to the original equation has the form z p ( x )= Ae x . Substituting this function into the given equation, we Fnd the constant A . z 0 + z = ( Ae x ) 0 + Ae x =2 Ae x =2 e x A =1 , and a general solution to the given nonhomogeneous equation is z ( x )= e x + c 1 cos x + c 2 sin x. Next, since z 0 ( x )= e x c 1 sin x + c 2 cos x , from the initial conditions we get a system for determining constants c 1 and c 2 . 0= z (0) = 1 + c 1 , 0= z 0 (0) = 1+ c 2 c 1 = 1 , c 2 =1 . Hence, z =( x )= e x cos x +sin x is the solution to the given initial value problem. 27. The roots of the auxiliary equation, r 2 r 2 = 0, are r = 1and r = 2. This gives a general solution to the corresponding homogeneous equation of the form y h ( x )= c 1 e x + c 2 e 2 x .W
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Unformatted text preview: use the superposition principle to Fnd a particular solution to the nonhomogeneous equation. (i) ±or the equation y − y − 2 y = cos x, a particular solution has the form y p, 1 ( x ) = A cos x + B sin x . Substitution into the above equation yields ( − A cos x − B sin x ) − ( − A sin x + B cos x ) − 2( A cos x + B sin x ) = ( − 3 A − B ) cos x + ( A − 3 B ) sin x = cos x ⇒ − 3 A − B = 1 , A − 3 B = 0 ⇒ A = − 3 / 10 , B = − 1 / 10 . So, y p, 1 ( x ) = − (3 / 10) cos x − (1 / 10) sin x . 201...
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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