# sol6 - r = 1 ± 2 i Thus y = t 1 2 i is a complex valued...

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MAP 2302 Fall-2011 Quiz 6 1. Find a general solution of the system x 0 + 2 y = 0 x 0 - y 0 = 0 SOLUTION. By subtracting the second equation from the ﬁrst we obtain y 0 + 2 y = 0 which has a general solution y = C 1 e - 2 t . Thus from the ﬁrst equation it follows that x 0 = - 2 C 1 e - 2 t . By integration we obtain x = C 1 e - 2 t + C 2 . The answer: x = C 1 e - 2 t + C 2 y = C 1 e - 2 t . 2. Find a general solution y 00 - 1 t y 0 + 5 t 2 y = 0 . SOLUTION. After multiplication by t 2 this turns into a Cauchy-Euler equation. For t > 0 we look for the fundamental solutions y 1 and y 2 of the form y = t r . The auxiliary equation is r 2 - 2 r + 5 = 0. It has complex roots
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Unformatted text preview: r = 1 ± 2 i . Thus, y = t 1+2 i is a complex valued solution. Note that t 1+2 i = tt 2 i = t ( e ln t ) 2 i = te (2 ln t ) i = t (cos(2 ln t ) + i sin(2 ln t )) . The real and the imaginary parts are linearly independent and hence they give the fundamental solution set y 1 = t cos(2 ln t ) and y 2 = t sin(2 ln t ). The general solution (fot t > 0) is y = C 1 y 1 + C 2 y 2 = C 1 t cos(2 ln t ) + C 2 t sin(2 ln t ) . 1...
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