Q5B_230 - Math 415 Quiz 5B The differential equation t2 y...

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Unformatted text preview: Math 415 Quiz 5B The differential equation t2 y + 5ty + 3y = 0 has y1 (t) = t−1 as one solution. Use the method of reduction of order to find a second solution. Since y1 (t) = t−1 is a solution, we assume that y2 is of the form v (t)y1 (t) for some unknown function v (t). Plugging this into the differential equation gives us 0 = t2 (v y1 + 2v y1 + vy1 ) + 5t(v y1 + vy1 ) + 3vy1 2 1 3 1 2 1 = t2 ( v − 2 v + 3 v ) + 5 t( v − 2 v ) + v t t t t t t = tv + 3v . So we have a first-order linear differential equation in v . For clarity, set w(t) = v (t). Then we have the differential equation tw + 3w = 0, or in the normalized form w + 3 w = 0. This has integrating factor t µ(t) = t3 , i.e. t3 w + 3t2 w = (t3 w) = 0, and so w(t) = D t3 . Which means that v (t) = Going back to our original assumption, we see that y2 (t) = v (t)y1 (t) = C1 t3 + C2 t. w(t) dt = C1 t4 + C2 . Since we only need one solution, which together with y1 forms a linearly independent pair, we can say that y2 (t) = t−3 . Find the solution of the initial value problem: y + y − 6y = −9te−t , y (0) = 0, y (0) = 1. The homogeneous equation y + y − 6y = 0 has the general solution C1 e2t + C2 e−3t . Note that e−t is not a solution to this version. That tells us that we can assume our specific solution is Y (t) = (At + B )e−t . Plugging this into the differential equation gives us −9te−t = (At + B − 2A)e−t + (−At + A − B )e−t − 6(At + B )e−t or −9t = −6At + (−A − 6B ). So −9 = −6A and 0 = −A − 6B , or A = 3/2 and B = −1/4 and the general solution will be y (t) = C1 e2t + C2 e−3t + 6t − 1 −t e. 4 Looking at our initial conditions tells us that 0 = C1 + C2 − 1 4 1 = 2C1 − 3C2 + 1 31 + 24 which we can put into standard form and then solve to find that C1 = 0 and C2 = 1/4. Making our final solution y (t) = 1 −3t (e + 6te−t − e−t ). 4 2 ...
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Q5B_230 - Math 415 Quiz 5B The differential equation t2 y...

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