185_pdfsam_math 54 differential equation solutions odd

185_pdfsam_math 54 differential equation solutions odd - 3...

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Exercises 4.3 25. The auxiliary equation, r 2 2 r + 2 = 0, has the roots r =1 ± i . Thus, a general solution is y ( t )= c 1 e t cos t + c 2 e t sin t, where c 1 and c 2 are arbitrary constants. To fnd the solution that satisfes the initial conditions, y ( π )= e π and y 0 ( π ) = 0, we fnd y 0 ( t )= c 1 e t (cos t sin t )+ c 2 e t (sin t +cos t ) and solve the system e π = y ( π )= c 1 e π , 0= y 0 ( π )= c 1 e π c 2 e π . This yields c 1 = 1, c 2 = c 1 =1. So,theansweris y ( t )= e t cos t + e t sin t = e t (sin t cos t ) . 27. To solve the auxiliary equation, r 3 4 r 2 +7 r 6 = 0, which is oF the third order, we fnd its real root frst. Examining the divisors oF 6, that is, ± 1, ± 2, ± 3, and ± 6, we fnd that r =2 satisfes the equation. Next, we divide r 3 4 r 2 +7 r 6by r 2andobta in r 3 4 r 2 +7 r 6=( r 2) ( r 2 2
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Unformatted text preview: + 3 ) . ThereFore, the other two roots oF the auxiliary equation are r = 2 ± √ 4 − 12 2 = 1 ± √ 2 i , and a general solution to the given di±erential equation is given by y ( t ) = c 1 e 2 t + c 2 e t cos √ 2 t + c 3 e t sin √ 2 t . Next, we fnd the derivatives, y ( t ) = 2 c 1 e 2 t + c 2 e t ± cos √ 2 t − √ 2 sin √ 2 t ² + c 3 e t ± sin √ 2 t + √ 2 cos √ 2 t ² , y ( t ) = 4 c 1 e 2 t + c 2 e t ± − cos √ 2 t − 2 √ 2 sin √ 2 t ² + c 3 e t ± − sin √ 2 t + 2 √ 2 cos √ 2 t ² , 181...
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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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