Exam 2, Solution 4 Jordan Paschke Problem 4. Suppose that y 1 (t) = t and y 2 (t) = t 3 are solutions to the equation 3t 3 y t 2 y 6ty + 6y = 0, t > 0. ( ) Find the general solution of the above equation. Solution. Since the equation is a third-order, linear, homogeneous differential equation, with particular solutions y 1 and y 2, then we know that its general solution will be of the form y(t) = c 1 y 1 (t) + c 2 y 2 (t) + c 3 y 3 (t) = c 1 t + c 2 t 3 + c 3 y 3 (t), where y 3 (t) is a third, linearly independent solution to ( ). method of reduction of order to help us find such a y 3. We will use the Reduction of order is used when we are trying to find the general solution to a homogeneous differential equation after having already found one or more particular solutions. To begin, we suppose that either y 3 (t) = y 1 (t)v(t) or y 3 (t) = y 2 (t)v(t), where v(t) will be some function to be determined later. Since it is easier to take multiple derivatives of t, rather than derivatives of t 3, we will work with y 1 = t. Supposing y 3 (t) = tv(t), we find its successive derivates and plug them into the original differential equation ( ): which gives us y 3 = tv y 3 = tv + v y 3 = tv + 2v y 3 = tv + 3v 1
or equivalently 3t 3 (tv + 3v ) t 2 (tv + 2v ) 6t(tv + v) + 6(tv) = 0 (3t 4 )v + (9t 3 t 3 )v + ( 2t 2 6t 2 )v + ( 6t + 6t)v = 0 (3t 4 )v + (8t 3 )v + ( 8t 2 )v = 0 v + 8 3t v + 8 3t 2 v = 0. ( ) Since this equation does not include any v terms (ones without any derivatives), then we may introduce a new function, w(t) = v (t), to rewrite the above equation as w + 8 3t w + 8 3t 2 w = 0. ( ) Notice that this is a second-order homogeneous equation, one less than the order of ( ), ergo the name reduction of order. Furthermore, if we can find the general solution of ( ), w(t), then we can find the general solution of ( ) by integrating, v(t) = w(t)dt. So let us recap what we have done so far: If y 3 (t) = tv(t) is a solution of ( ), then we must have that v(t) is a solution of ( ). Since ( ) can be rewritten in terms of w = v as ( ), then to find the general solution of v(t), it suffices to find the general solution of w(t) and integrate. This means that the in order to solve our original equation ( ), we need only to find the general solution of the equation ( ). Since this equation is a second-order, linear, homogeneous differential equation, then we know the general solution must be of the form w(t) = d 1 w 1 (t) + d 2 w 2 (t), where w 1 and w 2 form a fundamental set of solutions to ( ). That is w 1 and w 2 solve the differential equation and are linearly independent. Notice that if we could find one solution of ( ), say w 1 (t), then we could use reduction of order again to find a second solution. To find w 1 (t), we first look back at the original equation from the beginning of the problem, and notice that we may rewrite t 3 (a solution of the equation) as t(t 2 ). Plugging it into ( ) and simplifying as we did with v(t), we see that (t 2 ) + 8 3t (t2 ) + 8 3t 2 (t2 ) = 0. 2
That is to say, t 2 is a solution of ( ). Therefore (2t) + 8 3t (2t) + 8 (2t) = 0, 3t2 or, after factoring out and dividing by 2, that (t) + 8 3t (t) + 8 (t) = 0. 3t2 In other words, w 1 (t) = t is a solution of the homogeneous equation ( ). Therefore we may use reduction of order a second time to help us find another solution, w 2 (t) = w 1 (t)u(t) = tu(t). Using the same procedure as before we find: and or equivalently w 2 = tu w 2 = tu + u w 2 = tu + 2u (tu + 2u ) + 8 3t (tu + u) + 8 3t 2 (tu) = 0 (t)u + (2 + 8 3 )u + ( 8 3t 8 3t )u = 0 tu + 14 3 u = 0 u + 14 3t u = 0. ( ) Notice that this is again an equation involving no u terms, and so we may introduce the function z(t) = u (t) to rewrite it as z + 14 z = 0. ( ) 3t Finally we have arrived at a separable, first-order, linear differential equation which we may solve explicitly: z(t) = A 1 t 14/3. We can now work in reverse to finally recover the solution y 3 (t) we have been seeking all along. Since we have found the general solution for ( ), and since z = u, we may find the general solution for ( ) by integrating: u(t) = z(t)dt = A 1 t 14/3 dt = A 2 t 11/3 + A 3, where A 2 = 3 11 A 1. 3
From here, we wish to find a second solution to ( ), w 2 (t): w 2 (t) = w 1 (t)u(t) = t(a 2 t 11/3 + A 3 ) = A 2 t 8/3 + A 3 t. Since we only wish to find a second specific solution, linearly independent from w 1 (t) = t, we may take A 2 = 1 and A 3 = 0 to get w 2 (t) = t 8/3. This means that the general solution to ( ) is w(t) = d 1 t + d 2 t 8/3. To find the general solution of ( ) we note that w = v and integrate the above function to get v(t) = w(t)dt = d 1 t + d 2 t 8/3 dt = D 1 t 2 + D 2 t 5/3 + D 3, where D 1 = 1 2 d 1, D 2 = 3 5 d 2. Finally, we use this v(t) to find a solution of ( ), y 3 (t), that is linearly independent of y 1 (t) and y 2 (t): y 3 (t) = y 1 (t)v(t) = t(d 1 t 2 + D 2 t 5/3 + D 3 ) = D 1 t 3 + D 2 t 2/3 + D 3 t. Taking D 1 = D 3 = 0 and D 2 = 1, we find our desired solution y 3 (t) = t 2/3. Hence the final, general solution to the problem, is given by a linear combination of y 1, y 2, and y 3 : Shortcut: y(t) = c 1 t + c 2 t 3 + c 3 t 2/3. While the above method is a good exercise in the method of reduction of order, there is an interesting trick that can help you solve this differential equation much more quickly. Recall that in our study of linear differential equations with constant coefficients, we began by supposing that a solution was of the form e rt. From here, were took derivates and upon substituting into the differential equation, we arrived at a polynomial in r, the roots of which were the solutions we sought. 4
In this problem, suppose that a solution of ( ) is of the form y(t) = t k. We are already given that t and t 3 are solutions, so it s not unreasonable to assume a third solution would be of this form as well. Taking derivatives and plugging them into the equation, we find 3t 3 (k(k 1)(k 2)t k 3 ) t 2 (k(k 1)t k 2 ) 6t(kt k 1 ) + 6(t k ) = 0 3k(k 1)(k 2)t k k(k 1)t k 6kt k + 6t k = 0 [ 3k 3 10k 2 + k + 6 ] t k = 0 Hence, if k 0 is a root of p(k) = 3k 3 10k 2 + k + 6, then we must have that t k0 is a solution of ( ). In general, factoring p(k) might be difficult. However, since we already know that t 1 and t 3 are solutions of the equation, then we must have that k 1 = 1 and k 2 = 3 are roots of p(k). That is, we must have that both k 1 and k 3 divide p(k). After doing the polynomial long division, we are left with 3k + 2 = 0, which obviously has the solution k 3 = 2/3. Therefore y 3 (t) = t 2/3 must be a solution as well. 5