3-7-29 - 3.7.29 Consider the differential equation t2 y 2ty...

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Unformatted text preview: 3.7.29 Consider the differential equation t2 y - 2ty + 2y = 4t2 , t > 0. We are given that y1 (t) = t is a solution of the homogeneous problem (as we can verify by inspection). Thus by problem 28, y(t) = v(t)y1 (t) is also a solution, for v satisfying y1 v + [2y1 + p(t)y1 ]v = g(t), with p(t) = -2t/t2 = -2/t and g(t) = 4t2 /t2 = 4 in this case. Substituting our known functions into the above, we see that we are solving tv + [2 - 2]v = 4, (1) so v = 4/t, so v = 4 ln t + c1 . (We don't need absolute values in the log since t > 0.) Thus, recalling ln t dt = t ln t - t + c or by using integration by parts, we see v = 4(-t + t ln t) + c1 t + c2 = 4 ln t + c3 t + c2 where c3 = -4 + c1 . Thus our general solution is y = vy1 = 4t2 ln t + c3 t2 + c2 t. Usually our equation for v will not be as nice as in (1). Normally we would expect to make the substitution u = v , solve for u, then integrate again to find v, and then finally arrive at the general solution y = vy1 , as follows: Let u = v . Then we have tu + 0u = 4. Dividing both sides by y1 = t, we conclude u + 0 u = 4 . This is a first order linear ODE, and in this particular case the equation is t t separable, but worst-case-scenario we know we can always solve it by using an integrating factor (in this case, = 1). Then, 4 d (1u) = , dt t so u = 4 ln t + c4 , so v = udt = 4(-t + t ln t) + c4 t + c5 = 4t ln t + c6 t + c5 , so y = vy1 as above. Using an integrating factor is overkill for this particular problem, but it may be a good idea for 3.7.30. ...
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This note was uploaded on 07/17/2008 for the course MATH 415 taught by Professor Costin during the Fall '07 term at Ohio State.

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