258_pdfsam_math 54 differential equation solutions odd

# 258_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 ( A 6 B )cos3 t +(6 A + B )sin3 t =6cos3 t sin 3 t A 6 B =6 , 6 A + B = 1 A =0 , B = 1 . So, y p ( t )= sin 3 t ,and y ( t )=( c 1 cos 3 t + c 2 sin 3 t ) e t sin 3 t is a general solution to the given equation. Next, we satisfy the initial conditions. y (0) = c 1 =2 , y 0 (0) = c 1 +3 c 2 3= 8 c 1 =2 , c 2 = 7 / 3 . Hence, the answer is y ( t )= ± 2cos3 t 7 3 sin 3 t ² e t sin 3 t. 33. The associated characteristic equation in this problem is r 3 12 r 2 +27 r +40=0,whichis a third order equation. Using the rational root theorem, we look for its integer roots among the divisors of 40, which are ± 1, ± 2, ± 4, ± 8, ± 10, ± 20, and ± 40. By inspection, r = 1is a root. Dividing r 3 12 r 2 +27 r +40by r +1,weget r 3 12 r 2 +27 r +40=( r 2 13 r + 40)( r +1) , and so the other two roots of the auxiliary equation are the roots of r 2 13 r +40 = 0, which are r = 5 and 8. Therefore, a general solution to the given equation is y ( t )= c 1 e t + c 2 e 5 t + c 3 e 8 t . We Fnd the values of c 1 , c 2 ,and c 3 from the initial conditions.
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Unformatted text preview: y (0) = ( c 1 e − t + c 2 e 5 t + c 3 e 8 t ) ³ ³ t =0 = c 1 + c 2 + c 3 = − 3 , y (0) = ( − c 1 e − t + 5 c 2 e 5 t + 8 c 3 e 8 t ) ³ ³ t =0 = − c 1 + 5 c 2 + 8 c 3 = − 6 , y (0) = ( c 1 e − t + 25 c 2 e 5 t + 64 c 3 e 8 t ) ³ ³ t =0 = c 1 + 25 c 2 + 64 c 3 = − 12 ⇒ c 1 = − 1 , c 2 = − 3 , c 3 = 1 . Therefore, y ( t ) = − e − t − 3 e 5 t + e 8 t is the solution to the given initial value problem. 35. Since the roots of the auxiliary equation, r 2 + 1 = 0, are r = ± i , the functions y 1 ( θ ) = cos θ and y 2 ( θ ) = sin θ are two linearly independent solutions to the corresponding homogeneous equation, and its general solution is given by y h ( θ ) = c 1 cos θ + c 2 sin θ . 254...
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