{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

212_pdfsam_math 54 differential equation solutions odd

# 212_pdfsam_math 54 differential equation solutions odd -...

This preview shows page 1. Sign up to view the full content.

Chapter 4 Thus, a general solution to the equation describing the motion is y ( t ) = cos t + 1 2 sin t + c 1 e 3 t + c 2 e t . Differentiating, we find y ( t ) = sin t + (1 / 2) cos t 3 c 1 e 3 t c 2 e t . Initial conditions give y (0) = 1 + c 1 + c 2 = 1 / 2 , y (0) = 1 / 2 3 c 1 c 2 = 0 c 1 = 1 / 2 , c 2 = 2 . Hence, the equation of motion is y ( t ) = cos t + 1 2 sin t 1 2 e 3 t + 2 e t . 45. (a) With m = k = 1 and L = π given initial value problem becomes y ( t ) = 0 , t ≤ − π 2 V , y + y = cos V t, π/ (2 V ) < t < π/ (2 V ) , 0 , t π/ (2 V ) . The corresponding homogeneous equation y + y = 0 is the simple harmonic equation whose general solution is y h ( t ) = C 1 cos t + C 2 sin t . (4.2) First, we find the solution to the given problem for
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: V ) < t < π/ (2 V ). The nonhomogeneous term, cos V t , suggests a particular solution o± the ±orm y p ( t ) = A cos V t + B sin V t. Substituting y p ( t ) into the equation yields ( A cos V t + B sin V t ) + ( A cos V t + B sin V t ) = cos V t ⇒ ± − V 2 A cos V t − V 2 B sin V t ) + ( A cos V t + B sin V t ) = cos V t ⇒ ± 1 − V 2 ) A cos V t + ± 1 − V 2 ) B sin V t = cos V t . Equating coeﬃcients, we get A = 1 1 − V 2 , B = 0 , 208...
View Full Document

{[ snackBarMessage ]}