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solution_2.2_1

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

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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 μ ( t ) = e R p ( t ) dt = e R 3 dt = e 3 t . Hence the general solution is y ( t ) = R μ ( t ) g ( t ) dt + C μ ( 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 0 - 4 t y = 0 , from which we can see that p ( t ) = - 4 t , g ( t ) = 0 . The integrating factor is μ ( t ) = e R p ( t ) dt = e R - 4 /tdt = e - 4 ln | t | = e ln | t | - 4 = t - 4 . Hence that general solution is y ( t ) = R μ ( t ) g ( t ) dt + C μ ( 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

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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 0 - 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 . 12. For the given ODE, we have p ( t ) = 1 + sin ( t ) , g ( t ) = 0 . Z p ( t ) dt = Z (1 + sin ( t )) dt = t - cos ( t ) . The integrating factor is μ ( t ) = e R p ( t ) dt = e R (1+ sin ( t )) dt = e t - cos ( t ) .
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