Question

Using three terms in the Taylor series given in Eq. ( \(12)\) and taking \(h=0.1\), determine approximate values of the solution of the illustrative example \(y^{\prime}=1-t+4 y, y(0)=1\) at \(t=0.1\) and \(0.2 .\) Compare the results with those using the Euler method and with the exact values. Hint: If \(y^{\prime}=f(t, y),\) what is \(y^{\prime \prime \prime} ?\)

Short answer

Question: Approximate the values of \(y(0.1)\) and \(y(0.2)\) using Taylor series expansion with three terms for the given differential equation \(y'(t) = 1 - t + 4y(t)\) with initial condition \(y(0) = 1\). Also, compare the results with those from the Euler method. Answer: Using Taylor series expansion, the approximations for \(y(0.1)\) and \(y(0.2)\) are 1.595 and 2.38, respectively. In comparison, using the Euler method, we obtain approximations of 1.5 and 2.34 for the same values of \(t\).

Step-by-step solution

Find the second and third derivatives of \(y(t)\)

Given \(y'(t) = 1 - t + 4y(t)\). To find the second and third derivatives, we will differentiate both sides of the equation with respect to \(t\). For the second derivative: $$ y''(t) = \frac{d}{dt}(1-t+4y(t)) = -1 + 4y'(t) $$ For the third derivative: $$ y'''(t) = \frac{d}{dt}(-1+4y'(t)) = 4y''(t) $$

Write out the Taylor series expansion with three terms

The Taylor series expansion of \(y(t)\) with three terms is: $$ y(t) \approx y(0) + y'(0)t + \frac{1}{2}y''(0)t^2 $$

Plug in the initial condition and calculate the approximations for \(y(0.1)\) and \(y(0.2)\)

Now we will use the initial condition \(y(0) = 1\). We also need \(y'(0)\) and \(y''(0)\) which can be computed using the differential equation: $$ y'(0) = 1 - 0 + 4(1) = 5 \\ y''(0) = -1 + 4(5) = 19 $$ The Taylor series with three terms becomes: $$ y(t) \approx 1 + 5t + \frac{1}{2}(19)t^2 $$ Calculate the approximations for \(y(0.1)\) and \(y(0.2)\): $$ y(0.1) \approx 1 + 5(0.1) + \frac{1}{2}(19)(0.1)^2 = 1.595 \\ y(0.2) \approx 1 + 5(0.2) + \frac{1}{2}(19)(0.2)^2 = 2.38 $$

Apply the Euler method and compare the results

Let's apply the Euler method with a step size \(h = 0.1\): $$ y_{n+1} = y_n + h(1 - t_n + 4y_n) $$ For \(t = 0.1\), we have: $$ y_1 = y_0 + 0.1(1 - 0 + 4(1)) = 1.5 $$ For \(t = 0.2\), we have: $$ y_2 = y_1 + 0.1(1 - 0.1 + 4(1.5)) = 2.34 $$ Comparison of the approximations: | t | Taylor Series | Euler's Method | |------|--------------|----------------| | 0.1 | 1.595 | 1.5 | | 0.2 | 2.38 | 2.34 | The Taylor series approximation provides better results than the Euler's method, but more computational resources are needed to obtain higher-order derivatives.

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are fundamental in expressing physical laws, economic theories, and other phenomena where the rate of change is essential. A differential equation models a process that changes over time or space, like the growth of a population, the decay of radioactive material, or the changing speed of a falling object under gravity.

In the given exercise, the first-order differential equation is \(y' = 1 - t + 4y\), representing a dynamic system where the derivative \(y'\) of the unknown function \(y\) depends on the independent variable \(t\) and the function \(y\) itself. Solving such equations provides us with a function \(y(t)\) that predicts the system's state at any time \(t\).

Euler's Method

Euler's method is an introductory, numerical method for solving initial value problems for differential equations. It approximates solutions by taking a fixed step size to progress from an initial condition along the tangent of the curve that solves the differential equation. Though simple, Euler's method can be inaccurate over longer intervals or where the function's curvature is high.

In our exercise, starting from \(y(0) = 1\) and using a step size \(h = 0.1\), Euler's method gives us sequential values at \(t = 0.1\) and \(t = 0.2\) through the recurrence relation \(y_{n+1} = y_n + h(1 - t_n + 4y_n)\). This method is straightforward but does not always offer the precise refinement that methods relying on higher-order derivatives might.

Higher-Order Derivatives

Higher-order derivatives are the successive derivatives of a function, giving us deeper insight into the function's behavior. In the context of the Taylor series, these derivatives allow us to form more accurate approximations of the function by taking into account the curvature and how rapidly it changes.

In our exercise, the first derivative \(y'\) is given, and finding the second \(y''\) and third derivatives \(y'''\) involve differential calculus. \(y''\) is the acceleration of the change in \(y\), while \(y'''\) represents the 'jerk' or the rate of change of the acceleration. These derivatives contribute significantly to the precision of the function's Taylor series expansion.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation along with specific conditions to be satisfied by the solution at a given initial point. IVPs are crucial because they represent physical phenomena starting conditions. Solving an IVP means finding the function \(y(t)\) that not only satisfies the differential equation but also the initial conditions.

The equation \(y' = 1 - t + 4y\) with initial condition \(y(0) = 1\) poses an IVP where the function \(y(t)\) describing our system must pass through the point \( (0, 1) \). This initial condition allows us to solve for constant coefficients and makes it possible to approximate the solution curve starting from \(t = 0\).