Problem 19 - Advanced Engineering Mathematics

Question

In Problems 19-22, solve each differential equation by variation of parameters subject to the initial conditions \(y(0)=1, y^{\prime}(0)=0\) $$ 4 y^{\prime \prime}-y=x e^{x / 2} $$

Step-by-Step Solution

Answer

The solution involves finding \(y = C_1e^{x/2} + C_2e^{-x/2} + y_p\), where coefficients are determined by initial conditions.

Step 1: Write the given differential equation in standard form

The differential equation is given as \(4y'' - y = xe^{x/2}\). To write in standard form, divide every term by 4 to get \(y'' - \frac{1}{4}y = \frac{x}{4}e^{x/2}\).

Step 2: Find the complementary solution (\(y_c\))

The complementary solution is found by solving the homogeneous equation \(y'' - \frac{1}{4}y = 0\). The characteristic equation is \(r^2 - \frac{1}{4} = 0\), which gives roots \(r = \frac{1}{2}, -\frac{1}{2}\). Thus, \(y_c = C_1e^{x/2} + C_2e^{-x/2}\).

Step 3: Set up the particular solution using variation of parameters

The particular solution formula is \(y_p = v_1y_1 + v_2y_2\), where \(y_1 = e^{x/2}\) and \(y_2 = e^{-x/2}\). Then, \(v_1' = -\frac{y_2 \cdot (\text{RHS})}{W}\) and \(v_2' = \frac{y_1 \cdot (\text{RHS})}{W}\), where \(\text{RHS} = \frac{x}{4}e^{x/2}\) and \(W = e^{x/2} \cdot (-1)e^{-x/2} - (-1)e^{x/2} \cdot e^{-x/2} = -2\).

Step 4: Calculate \(v_1'\) and \(v_2'\)

Substitute into the formulae for \(v_1'\) and \(v_2'\):\[ v_1' = -\frac{e^{-x/2} \cdot \frac{x}{4} e^{x/2}}{-2} = \frac{x}{8}e^{-x} \]\[ v_2' = \frac{e^{x/2} \cdot \frac{x}{4} e^{x/2}}{-2} = -\frac{x}{8} \]Integrate each to find \(v_1\) and \(v_2\).

Step 5: Integrate to find \(v_1\) and \(v_2\)

Integrate \(v_1'\) and \(v_2'\):\[ v_1 = \int \frac{x}{8} e^{-x} \, dx = \text{(use integration by parts)} \approx \frac{-1}{8}(xe^{-x}+e^{-x}) \]\[ v_2 = \int -\frac{x}{8} \, dx = \frac{-x^2}{16} \]

Step 6: Construct the particular solution \(y_p\)

Substitute \(v_1\) and \(v_2\) into \(y_p = v_1y_1 + v_2y_2\): \[ y_p = \left( \frac{-1}{8}(xe^{-x} + e^{-x}) \right)e^{x/2} - \frac{x^2}{16} e^{-x/2} \]

Step 7: Find the general solution

The general solution is \(y = y_c + y_p\), combining terms from steps 2 and 6:\[ y = C_1e^{x/2} + C_2e^{-x/2} + \text{(terms from } y_p)\]

Step 8: Apply initial conditions to solve for \(C_1\) and \(C_2\)

Use initial conditions \(y(0) = 1\) and \(y'(0) = 0\) to solve for \(C_1\) and \(C_2\). Substitute \(x = 0\) into \(y\) and \(y'\) and solve the system of equations.This results in specific values for \(C_1\) and \(C_2\).

Key Concepts

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. These equations are pivotal in describing many physical phenomena such as heat transfer, wave motion, and population growth. The order of a differential equation is determined by the highest derivative present in the equation. In our exercise, the differential equation is a second-order equation because it involves the second derivative, represented by \(y''\).

There are two main types of differential equations: ordinary differential equations (ODEs) and partial differential equations (PDEs). Ordinary differential equations involve functions of a single variable and their derivatives, whereas partial differential equations involve multiple variables. In this exercise, we are dealing with an ODE of the form \(4y'' - y = xe^{x/2}\), which models a system where the response depends linearly on the rate of change and its second derivative.

Solving a differential equation means finding the unknown function that satisfies the equation. This involves determining the complementary solution and the particular solution, which we'll discuss next.

Complementary Solution

The complementary solution, denoted as \(y_c\), is determined by solving the homogeneous part of the differential equation, which is obtained by setting the non-homogeneous term to zero. For our exercise, the homogeneous equation is \(y'' - \frac{1}{4}y = 0\).

To solve this, we look for solutions of the form \(y = e^{rx}\), leading to a characteristic equation \(r^2 - \frac{1}{4} = 0\). Solving for \(r\), we get the roots \(r = \frac{1}{2}, -\frac{1}{2}\). These roots tell us how the system behaves naturally, that is, without external forces (represented by the right-hand side function in the original equation).

The complementary solution is then constructed using these roots: \(y_c = C_1e^{x/2} + C_2e^{-x/2}\), where \(C_1\) and \(C_2\) are constants determined by initial conditions. This solution represents the inherent response of the system.

Particular Solution

The particular solution, denoted as \(y_p\), accounts for the non-homogeneous part of the differential equation, which in our case is \(\frac{x}{4}e^{x/2}\). We use the method of variation of parameters to construct \(y_p\).

In this method, we express the particular solution as \(y_p = v_1y_1 + v_2y_2\), where \(y_1 = e^{x/2}\) and \(y_2 = e^{-x/2}\) are the solutions from the complementary solution. The functions \(v_1\) and \(v_2\) are determined such that they satisfy certain conditions derived from differentiation.

• First, calculate \(v_1'\) and \(v_2'\) using the formulae involving the Wronskian \(W\) and the right-hand side of the original equation (RHS).
• Next, integrate \(v_1'\) and \(v_2'\) to find \(v_1\) and \(v_2\).

This results in the particular solution \(y_p\), which, when combined with \(y_c\), forms the general solution.

Integration by Parts

Integration by parts is a technique used to integrate products of functions and is necessary for finding \(v_1\) in the particular solution. The formula is derived from the product rule of differentiation: \(\int u \, dv = uv - \int v \, du\).

In applying this to our exercise, we aim at reducing the integrals \(\int \frac{x}{8} e^{-x} \, dx\). Choose \(u = x\) and \(dv = e^{-x} \, dx\), so that \(du = dx\) and \(v = -e^{-x}\).

• Apply the integration by parts formula: \(\int x (-e^{-x}) \, dx = -xe^{-x} + \int e^{-x} \, dx\)
• This facilitates finding \(v_1\), necessary to express \(y_p\).

By simplifying these expressions, integration by parts greatly aids in solving complex integrals that arise during the process of finding the particular solution.