Solving The Differential Equation D²s/dt² = 8t With Initial Conditions

In this article, we delve into the realm of differential equations, specifically focusing on solving the second-order differential equation d²s/dt² = 8t. This equation describes the relationship between the second derivative of a function s with respect to time t and the time variable itself. To obtain a unique solution for s(t), we are provided with initial conditions: at time t = 0, the function s has a value of 5 (s(0) = 5), and its first derivative ds/dt has a value of 4 ((ds/dt)(0) = 4). This kind of problem often arises in physics, engineering, and other fields when modeling the motion of objects, the behavior of circuits, or the dynamics of systems. The process of solving this differential equation involves integrating twice to eliminate the derivatives and then using the initial conditions to determine the integration constants. Differential equations are powerful mathematical tools used to model a wide variety of phenomena in science and engineering. They describe the relationship between a function and its derivatives, allowing us to understand and predict how systems change over time. Solving differential equations often involves techniques such as integration, separation of variables, and the use of initial conditions or boundary conditions. In this particular problem, we are presented with a second-order differential equation, which means that the highest derivative present is the second derivative. Solving this type of equation typically requires two integrations, each introducing an arbitrary constant of integration. These constants are then determined using the given initial conditions, which provide specific values of the function and its first derivative at a particular point in time. The initial conditions act as anchors, ensuring that we obtain a unique solution that satisfies both the differential equation and the specified values at the initial time. Without these conditions, we would have a family of solutions, each differing by a constant term. This problem serves as an excellent example of how mathematical techniques can be applied to solve real-world problems, providing a deeper understanding of the underlying processes and allowing for accurate predictions of future behavior.

Step 1: First Integration

Our first step in tackling this problem is to integrate both sides of the equation d²s/dt² = 8t with respect to t. This integration will reduce the order of the derivative by one. The left-hand side, the integral of d²s/dt², becomes ds/dt, which represents the first derivative of s with respect to t. The right-hand side, the integral of 8t, becomes 4t² + C₁, where C₁ is the first constant of integration. Therefore, after the first integration, we obtain the equation ds/dt = 4t² + C₁. This equation tells us how the rate of change of s varies with time. The constant C₁ represents an initial velocity or rate of change, which we will determine using the given initial condition. To determine the value of C₁, we use the initial condition (ds/dt)(0) = 4. This condition states that at time t = 0, the first derivative ds/dt has a value of 4. Substituting t = 0 and ds/dt = 4 into the equation ds/dt = 4t² + C₁, we get 4 = 4(0)² + C₁, which simplifies to C₁ = 4. This means that the initial rate of change of s is 4 units per unit of time. Now that we have found the value of C₁, we can rewrite the equation for ds/dt as ds/dt = 4t² + 4. This equation completely describes the first derivative of s with respect to t, taking into account both the time-dependent term 4t² and the constant initial rate of change 4. The first integration is a crucial step in solving the differential equation, as it reduces the order of the equation and allows us to move closer to finding the solution for s(t). By introducing the constant of integration C₁ and using the initial condition, we have successfully determined the value of this constant and obtained a more specific equation for the first derivative. This equation will be the starting point for the second integration, which will ultimately lead us to the solution for s(t).

Step 2: Second Integration

Having found the first derivative, our next step is to integrate the equation ds/dt = 4t² + 4 once again with respect to t. This second integration will give us the function s(t) itself. Integrating the left-hand side, ds/dt, with respect to t yields s(t). Integrating the right-hand side, 4t² + 4, with respect to t results in (4/3)t³ + 4t + C₂, where C₂ is the second constant of integration. Therefore, after the second integration, we obtain the equation s(t) = (4/3)t³ + 4t + C₂. This equation represents the general solution to the given differential equation. It describes a family of functions that satisfy the equation, each differing by the constant term C₂. To find the specific solution that satisfies the given initial conditions, we need to determine the value of C₂. To determine the value of C₂, we use the initial condition s(0) = 5. This condition states that at time t = 0, the function s(t) has a value of 5. Substituting t = 0 and s(0) = 5 into the equation s(t) = (4/3)t³ + 4t + C₂, we get 5 = (4/3)(0)³ + 4(0) + C₂, which simplifies to C₂ = 5. This means that the constant term in the specific solution is 5. Now that we have found the value of C₂, we can write the complete solution for s(t) as s(t) = (4/3)t³ + 4t + 5. This equation represents the unique function that satisfies both the differential equation d²s/dt² = 8t and the initial conditions s(0) = 5 and (ds/dt)(0) = 4. The second integration is the final step in solving the differential equation. By integrating the first derivative and using the second initial condition, we have successfully determined the function s(t) that describes the relationship between s and t. This solution provides a complete understanding of the behavior of the system being modeled by the differential equation.

Step 3: The Solution

After performing the two integrations and applying the initial conditions, we arrive at the specific solution for the function s(t). The solution is given by the equation s(t) = (4/3)t³ + 4t + 5. This equation represents a cubic function of time t. The term (4/3)t³ indicates that the function s(t) will exhibit a cubic growth pattern as time increases. This means that the rate of change of s(t) will increase at an accelerating rate. The term 4t represents a linear growth component. This part of the function contributes to a steady increase in s(t) over time. The constant term 5 represents the initial value of the function s(t) at time t = 0. This is the value that was given in the initial condition s(0) = 5. The solution s(t) = (4/3)t³ + 4t + 5 completely describes the behavior of s as a function of time t. It satisfies both the original differential equation d²s/dt² = 8t and the initial conditions s(0) = 5 and (ds/dt)(0) = 4. This solution can be used to predict the value of s at any given time t. It provides a complete understanding of the relationship between s and t in the context of the problem. In summary, the solution s(t) = (4/3)t³ + 4t + 5 is the culmination of the integration process and the application of initial conditions. It provides a precise and accurate representation of the function s(t), allowing us to analyze and predict its behavior over time. This solution demonstrates the power of differential equations in modeling and solving problems in various fields of science and engineering.

Conclusion

In conclusion, we have successfully solved the second-order differential equation d²s/dt² = 8t with the initial conditions s(0) = 5 and (ds/dt)(0) = 4. The process involved two integrations, each introducing a constant of integration. These constants were determined using the initial conditions, leading to the unique solution s(t) = (4/3)t³ + 4t + 5. This solution represents the function s(t) that satisfies both the differential equation and the initial conditions. It provides a complete description of the relationship between s and t in the context of the problem. The solution demonstrates the power of differential equations in modeling and solving problems in various fields of science and engineering. The steps involved in solving this problem are applicable to a wide range of differential equations. First, integrate the equation to reduce the order of the derivative. Second, use the initial conditions to determine the constants of integration. Finally, write the complete solution, which represents the function that satisfies both the differential equation and the initial conditions. The solution s(t) = (4/3)t³ + 4t + 5 is a cubic function of time t. This means that the function s(t) will exhibit a cubic growth pattern as time increases. The cubic term (4/3)t³ dominates the behavior of the function for large values of t. The linear term 4t contributes to a steady increase in s(t) over time. The constant term 5 represents the initial value of the function s(t) at time t = 0. This solution can be used to predict the value of s at any given time t. It provides a complete understanding of the relationship between s and t in the context of the problem. Overall, this exercise highlights the importance of differential equations and initial conditions in modeling and solving real-world problems. The techniques used in this solution can be applied to a wide range of problems in physics, engineering, and other fields.