HW7 - MA341 Homework 7 Solution 4.6 5 y 00 2 y y = t 1 e t The characteristic equation for the corresponding homogeneous equation is r 2 2 r 1 = 0

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Unformatted text preview: MA341 Homework 7 Solution 4.6 5. y 00- 2 y + y = t- 1 e t . The characteristic equation for the corresponding homogeneous equation is r 2- 2 r + 1 = 0; it has a double root: r = 1. Thus, e t and te t are two linearly independent solutions to the homogeneous equation. Then, we take y p ( t ) = v 1 ( t ) e t + v 2 ( t ) te t , and, referring to formula (9) in section 4.6, solve the system v 1 ( t ) e t + v 2 ( t ) te t = 0 , v 1 ( t ) e t + v 2 ( t )( e t + te t ) = t- 1 e t , for v 1 ( t ) and v 2 ( t ). This yields: v 1 ( t ) =- 1 , v 2 ( t ) = t- 1 . Integrating, we obtain: v 1 ( t ) = Z- 1 dt =- t + C 1 , v 2 ( t ) = Z t- 1 dt = ln | t | + C 2 . We need only one particular solution, so we take both C 1 and C 2 to be zero for simplicity. Then, substituting v 1 ( t ) and v 2 ( t ) into y p ( t ), we obtain: y p ( t ) =- te t + ln | t | te t . The first term on the right-hand side of the previous equality is a part of the general solution to the homogeneous equation. Consequently, a general solution to the given inhomogeneous equation can be taken in the form: y ( t ) = C 1 e t + C 2 te t + ln | t | te t . 9. y 00 + 4 y = csc 2 (2 t ) . The characteristic equation for the corresponding homogeneous equation is r 2 +4 = 0; it has two complex-conjugate roots: r = ± 2 i . Thus, cos 2 t and sin 2 t are two linearly independent solutions to the homogeneous equation. Then, we take y p ( t ) = v 1 ( t ) cos 2 t + v 2 ( t ) sin 2 t, and, referring to formula (9) in section 4.6, solve the system (cos 2 t ) v 1 ( t ) + (sin 2 t ) v 2 ( t ) = 0 , (- 2 sin 2 t ) v 1 ( t ) + (2 cos 2 t ) v 2 ( t ) = csc 2 (2 t ) , for v 1 ( t ) and v 2 ( t ). This yields: v 1 ( t ) =- 1 2 csc(2 t ) , 1 MA341 Homework 7 Solution v 2 ( t ) = 1 2 cos 2 t csc 2 (2 t ) . Integrating, we obtain: v 1 ( t ) = Z- 1 2 csc(2 t ) dt = 1 4 ln | csc(2 t ) + cot(2 t ) | + C 1 , v 2 ( t ) = Z 1 2 cos 2 t csc 2 (2 t ) dt = 1 4 Z 1 sin 2 (2 t ) d (sin 2 t ) =- 1 4 csc(2 t ) + C 2 . We need only one particular solution, so we take both C 1 and C 2 to be zero for simplicity. Then, substituting v 1 ( t ) and v 2 ( t ) into y p ( t ), we obtain: y p ( t ) = 1 4 [(cos 2 t ) ln | csc(2 t ) + cot (2 t ) | - 1] . Consequently, a general solution to the given equation can be taken in the form: y ( t ) = C 1 cos 2 t + C 2 sin 2 t + 1 4 [(cos 2 t ) ln | csc(2 t ) + cot(2 t ) | - 1] . 11. y 00 + y = tan 2 ( t ) The characteristic equation for the corresponding homogeneous equation is r 2 + 1 = 0; it has two complex-conjugate roots: r = ± i . Thus, cos t and sin t are two linearly independent solutions to the homogeneous equation. Then, we take y p ( t ) = v 1 ( t ) cos t + v 2 ( t ) sin t, and, referring to formula (9) in section 4.6, solve the system (cos t ) v 1 ( t ) + (sin t ) v 2 ( t ) = 0 , (- sin t ) v 1 ( t ) + (cos t ) v 2 ( t ) = tan 2 t, for v 1 ( t ) and v 2 ( t ). This yields: v 1 ( t ) =- sin 3 t cos 2 t , v 2 ( t ) = sin 2 t cos t ....
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This homework help was uploaded on 04/16/2008 for the course MA 341 taught by Professor Schecter during the Spring '08 term at N.C. State.

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HW7 - MA341 Homework 7 Solution 4.6 5 y 00 2 y y = t 1 e t The characteristic equation for the corresponding homogeneous equation is r 2 2 r 1 = 0

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