D solve the differential equation x 2 y 4 xy 14 y 0

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d. Solve the differential equation x 2 y ′′ - 4 xy - 14 y = 0. Hint: the two linearly independent solutions are of the form y = x r for appropriate values of r . Solution: We are told to try y = x r , so we do precisely that: y = x r , y = rx r 1 , y ′′ = r ( r - 1) x r 2 . Making these substitutions, we get r ( r - 1) x r - 4 rx r - 14 x r = 0 , which simplifies to the equation r 2 - 5 r - 14 = 0 ,
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MAP 2302, Fall 2011 — Final Exam Review Problems 13 which has solutions r = - 2 and r = 7. This shows us that the solution is y ( x ) = Cx 2 + Dx 7 . e. A 1 kg mass is attached to a spring with stiffness constant k = 9 N/m. The damping constant for the system is b = 6 Ns/m. If the mass is pushed rightward from the equilibrium position with an initial velocity of 2 m/s, what will its maximum displacement be? Solution: We first need to determine the constants for the free spring equation, my ′′ + by + ky = 0 . In this case they are basically given to us, and the units all match, so we quickly get y ′′ + 6 y + 9 y = 0 . The characteristic polynomial is r 2 + 6 r + 9 = ( r + 3) 2 , which has the repeated real root r = - 3 (so, the system is critically damped ). Hence the solution is y = Ce 3 t + Dte 3 t for appropriate constants C and D . Now we must determine our initial conditions. We are given that the mass “is pushed rightward from the equilibrium position with an initial velocity of 2 m/s”. This implies that it starts at the equilibrium position, so y (0) = 0. Assuming that positive y values correspond to distance to the right of equilibrium, we take y (0) = 2. First we plug in t = 0: y (0) = Ce 0 + D 0 e 0 = C = 0 , so C = 0, and thus y = Dte 3 t . Taking a derivative, we see that y = De 3 t - 3 Dte 3 t . Plugging t = 0 into this equation, we get that y (0) = De 0 - 3 D 0 e 0 = D = 2 , so D = 2. We now have the complete solution to this initial value problem, y ( t ) = 2 te 3 t .
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MAP 2302, Fall 2011 — Final Exam Review Problems 14 Now we just need to compute the maximum displacement. We know that this will occur either at t = 0 or at the first critical point. However, it cannot occur when t = 0, because y (0) = 0, so it must happen at the first critical point. Setting the first derivative equal to 0 we have y = 2 e 3 t - 6 te 3 t = 0 . Now we can divide through by e 3 t to see that 2 - 6 t = 0, or equivalently, t = 1 / 3 . Therefore the maximum displacement is y ( 1 / 3 ) = 2 3 e 1 .
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MAP 2302, Fall 2011 — Final Exam Review Problems 15 Formulas of possible use sin 2 x = 2 sin x cos x L { t n } = n ! s n +1 cos 2 x = cos 2 x - sin 2 x L braceleftbig e at bracerightbig = 1 s - a sin 2 x = 1 - cos 2 x 2 L { sin bt } = b s 2 + b 2 cos 2 x = 1 + cos 2 x 2 L { cos bt } = s s 2 + b 2 sin( x + y ) = sin x cos y + cos x sin y L braceleftbig e at f ( t ) bracerightbig = F ( s - a ) , where F ( s ) = L { f } . cos( x + y ) = cos x cos y - sin x sin y L { tf ( t ) } = - d ds L { f } integraldisplay tan u du = - ln | cos u | L braceleftbig f ( y ) bracerightbig = s L { f } - f (0) integraldisplay cot u du = ln | sin u | L { u ( t - a ) f ( t ) } = e as L { f ( t + a ) } integraldisplay sec u du = ln | sec u + tan u | L { f , period T } = L { f T } 1 - e sT integraldisplay csc u du = ln | csc u - cot u | , L { δ ( t - a ) } = e as sin x = summationdisplay n =0 ( - 1) n x 2 n +1 (2 n + 1)!
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