solution_2.2_1

solution_2.2_1 - Solutions to selected problems in Section...

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Unformatted text preview: Solutions to selected problems in Section 2.2 1. (a). Note that for the give ODE, we have p ( t ) = 3 ,g ( t ) = 0 . The integrating factor is ( t ) = e R p ( t ) dt = e R 3 dt = e 3 t . Hence the general solution is y ( t ) = R ( t ) g ( t ) dt + C ( t ) = Ce- 3 t . (b). By the initial condition: C = y (0) =- 3 , and the solution to the IVP is y ( t ) =- 3 e- 3 t . 4. (a). We first need to reduce the given ODE to the standard form: y- 4 t y = 0 , from which we can see that p ( t ) =- 4 t ,g ( t ) = 0 . The integrating factor is ( t ) = e R p ( t ) dt = e R- 4 /tdt = e- 4ln | t | = e ln | t |- 4 = t- 4 . Hence that general solution is y ( t ) = R ( t ) g ( t ) dt + C ( t ) = Ct 4 . (b). By the initial condition, C 1 4 = y (1) = 1 , we have C = 1 and the solution to the IVP is y ( t ) = t 4 . 1 6. Since p ( t ) =- 2, the integrating factor is e- 2 t . Hence we can put the equation in the form e- 2 t ( y- 2 y ) = d dt ( e- 2 t y ) = e t . Integration leads to e- 2 t y = e t + C, i.e. y = e 3 t + Ce 2 t . Finally the initial condition yields y (0) = 1 + C = 3 , i.e. C = 2, and the solution to this IVP is y ( t ) = e 3 t + 2 e 2 t ....
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This note was uploaded on 10/16/2009 for the course MATH 2214 at Virginia Tech.

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solution_2.2_1 - Solutions to selected problems in Section...

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