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2. Solve the initial value problem. 11(t + 1) dy − 9y = 18t, dx for t > 0 with y(0) = 6. Solution: Dividing both sides by 11(t + 1), we rewrite the differential equation as follows: 18 t 9 y0 − y= (1) 11(t + 1) 11 t + 1 so we can identify the coefficient P (t) = − 9 . 11(t + 1) We then find the integrating factor, I(t) = e R P (t) dt 9 = e− 11 R 1 t+1 dt 9 = e− 11 ln(t+1) = (t + 1)−9/11 . Multiply both sides of equation (1) by I(t) to get (t + 1)−9/11 y 0 − 18 9 (t + 1)−9/11−1 y = (t + 1)−9/11 (t + 1)−1 t. 11 11 Notice the left hand side is the derivative of the product of I(t) and y. So, simplifying, we have 0 18 t (t + 1)−9/11 y = . 11 (t + 1)20/11 Now integrate both sides. On the left, this is simply (t+1)−9/11 y. You integrate the right-hand side as I showed you in class. So, we have, 11 18 11 2/11 −9/11 −9/11 (t + 1) + (t + 1) +C (t + 1) y= 11 2 9 = 9(t + 1)2/11 + 2(t + 1)−9/11 + C. Now, here is where I made a mistake in class. When you solve for y, be careful to multiply everything on the right-hand side by (t + 1)9/11 , even C, as follows: y = 9(t + 1)9/11 (t + 1)2/11 + 2(t + 1)9/11 (t + 1)−9/11 + (t + 1)9/11 C = 9(t + 1) + 2 + (t + 1)9/11 C. (2) Finally, you must find C so that y(0) = 6. That is, 6 = y(0) = 9(0 + 1) + 2 + (0 + 1)9/11 C = 11 + C. So, C = −5, (as we correctly computed in class). Plug this value for C into equation (2) to get the final correct solution: y = 9(t + 1) + 2 − 5(t + 1)9/11 . 1