Q1_130 - Math 415 Name: Quiz 1A Verify that y1 (t) = t/4...

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Math 415 Name: Quiz 1A Verify that y 1 ( t ) = t/ 4 and y 2 ( t ) = e - t + t/ 4 are solutions to y 0000 + 5 y 000 + 4 y = t . We first take derivatives: y 0 1 ( t ) = 1 / 4 so y 00 1 ( t ) = 0 = y 000 1 ( t ) = y 0000 1 ( t ). Also, y 0 2 ( t ) = - e - t + 1 / 4, y 00 2 ( t ) = e - t , y 000 2 ( t ) = - e - t , and y 0000 2 ( t ) = e - t . So y 0000 1 + 5 y 000 1 + 4 y 1 = 0 + 0 + 4( t/ 4) = t and y 0000 2 + 5 y 000 2 + 4 y 2 = e - t - 5 e - t + 4( e - t + t/ 4) = t Find the solution to y 0 - y = 2 te 3 t , y (0) = 2. If we multiply through by an integrating factor, we get μy 0 - μy = 2 tμe 3 t . For the left side to be the result of applying the product rule we must have dt = - μ which is a separable equation with solution μ ( t ) = e - t . This makes the differential equation ( e - t y ) 0 = 2 te 2 t . The right side has integral te 2 t - 1 2 e 2 t + C . Plugging in the initial condition tells us that e - 0 · 2 = 0 · e 2 · 0 - 1 2 e 2 · 0 + C , or C = 5 / 2. Our final solution is y ( t ) = te 3 t - 1 2 e 3 t + 5 2 e t . 1
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Math 415 Name: Quiz 1B Show that ( A~x,~ y ) = ( ~x,A * ~ y ). Hint: Note that the dot product ( ~a,
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Q1_130 - Math 415 Name: Quiz 1A Verify that y1 (t) = t/4...

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