Final answer:
The solution to the IVP Gy"' + y' = 0 with given initial conditions is y(t) = -2*cos(t) + sin(t) - 1, which oscillates indefinitely as t approaches infinity. The trigonometric nature of the solution ensures that it neither grows nor decays unboundedly over time.
Step-by-step explanation:
To find the solution of the given initial value problem (IVP) with the differential equation Gy"' + y' = 0 and the initial conditions y (0) = 1, y' (0) = 1, y" (0) = 2, we will follow a step-by-step approach. Firstly, since the highest order derivative is y"', we require three integration constants that will be determined using the initial conditions.
Step 1: Integrate the equation Gy" + y = C1 once with respect to t to get y" + G^{-1}y = G^{-1}C1. Let's assume G = 1 for simplicity.
Step 2: The general solution of the second-order differential equation y" + y = C1 is y(t) = C2*cos(t) + C3*sin(t) + C1.
Step 3: Apply the initial conditions to find C1, C2, and C3. Using y (0) = 1 gives C2 + C1 = 1; y' (0) = 1 gives C3 = 1, and y" (0) = 2 gives -C2 = 2. This leads to C1 = -1, C2 = -2, and C3 = 1.
Step 4: The specific solution for this IVP is y(t) = -2*cos(t) + sin(t) - 1.
As t→∞, the trigonometric terms oscillate between -1 and 1, which becomes negligible as compared to any term that might grow with t. However, since our solution contains only trigonometric terms and a constant, the solution will continue to oscillate indefinitely with no unbounded increase or decrease in amplitude. Hence, the solution exhibits an oscillatory behavior as t→∞.
To plot its graph, you would mark the initial conditions on the axes and sketch sinusoidal curves that fit these conditions, oscillating around y = -1. Remember, accuracy in sketching the graph is key to visualizing the periodic behavior of the solution with time.