Question

Use variation of parameters to solve the initial value problem and graph the solution, given that \(y_{1}, y_{2}\) are solutions of the complementary equation. $$ \begin{array}{ll} \text (x+1)(2 x+3) y^{\prime \prime}+2(x+2) y^{\prime}-2 y=(2 x+3)^{2}, \quad y(0)=0, & y^{\prime}(0)=0 ; \\ y_{1}=x+2, \quad y_{2}=\frac{1}{x+1} \end{array} $$

Short answer

#Answer#: The solution to the initial value problem is \(y(x) = \frac{x^2}{(x + 1)^2} - \frac{x}{x + 1} - \frac{1}{x+1}\) with the initial conditions \(y(0) = 0\) and \(y'(0) = 0\).

Step-by-step solution

Apply the method of variation of parameters

To apply the method of variation of parameters, we first write the particular solution in the form \(y_p = u_1 y_1 + u_2 y_2\), where \(u_1\) and \(u_2\) are functions of \(x\) to be determined. We need to find \(u_1'\) and \(u_2'\) first and then integrate them to find \(u_1\) and \(u_2\). We have the system of equations: $$u_1' y_1 + u_2' y_2 = 0$$ $$u_1' y_1' + u_2' y_2' = g(x)$$ where \(g(x) = \frac{(2x+3)^2}{(x+1)(2x+3)}\).

Solve for Wronskian

We find the Wronskian \(W(y_1, y_2)\) of \(y_1\) and \(y_2\): $$W(y_1,y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} x+2 & \frac{1}{x+1} \\ 1 & \frac{-1}{(x+1)^2} \end{vmatrix} = -\frac{1}{(x+1)^3} - \frac{(x + 2)}{(x+1)^2}$$

Use the Wronskian to find particular solution

Next, we find the particular solution using the variation of parameters formula: $$u_1'= \frac{-y_2 g(x)}{W(y_1, y_2)}$$ $$u_2'= \frac{y_1 g(x)}{W(y_1, y_2)}$$ Evaluate each expression and integrate both sides. $$ u_1(x) = \int \frac{-y_2 g(x)}{W(y_1, y_2)} dx = - \int \frac{(2x+3)}{(x+1)(2x+3)^2} dx $$ $$ u_2(x) = \int \frac{y_1 g(x)}{W(y_1, y_2)} dx = \int \frac{(x+2)(2x+3)}{(x+1)^4} dx $$ After solving these integrals, we get: $$u_1(x)=-\frac{1}{(x+1)^2}+c_1$$ $$u_2(x)=-\frac{x^{2}}{4(x+1)^{2}}+\frac{x}{2(x+1)}-1+c_2$$

Apply initial conditions

Now we have that \(y_p = u_1 y_1 + u_2 y_2\). We can apply the initial conditions as follows: 1. \(y(0) = 0 = u_1(0)y_1(0) + u_2(0)y_2(0) \Rightarrow -1+2c_1 + c_2 = 0\) 2. \(y'(0) = 0 = u_1(0)y_1'(0) + u_2(0)y_2'(0) \Rightarrow -2 + \frac{c_2}{-1} = 0\) By solving these two equations, we get that \(c_1 = \frac{1}{2}\) and \(c_2 = -2\).

Obtain the complete solution

Plug these values back into our expressions for \(u_1\) and \(u_2\), and substitute them back into the particular solution. $$y_p(x) = \left(-\frac{1}{(x+1)^2} + \frac{1}{2}\right)(x+2) +\left(-\frac{x^2}{4(x+1)^2} + \frac{x}{2(x+1)} - 1 - 2\right)\left(\frac{1}{x+1}\right)$$ The final solution is: $$y(x) = y_p(x) = \frac{x^2}{(x + 1)^2} - \frac{x}{x + 1} - \frac{1}{x+1}$$

Graph the solution

Now that we have found the final solution for the given initial value problem, it's time to graph it. To do so, we can use any graphing tool to plot the function \(y(x) = \frac{x^2}{(x+1)^2} - \frac{x}{x+1} - \frac{1}{x+1}\), and examine the behavior of the function, ensuring it abides by the initial conditions given.

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific initial conditions. It provides the starting values for the function and its derivatives. This ensures a unique solution to the differential equation rather than the general solution. In the exercise, the initial value problem is defined as: \( y(0) = 0 \), \( y'(0) = 0 \). These conditions specify the value of the function \(y\) and its derivative \(y'\) at \(x = 0\). Knowing these values is crucial because they guide the resolution process of the differential equation, ensuring that the resulting solution meets these initial specifications.

Complementary Equation

A complementary equation is derived from the homogeneous part of the differential equation. It allows us to understand the general behavior of the solution without external forces or inputs. In this exercise, the complementary equation is derived by omitting the non-homogeneous term, \((2x+3)^2\). The solutions\(y_1 = x + 2\) and \(y_2 = \frac{1}{x + 1}\) are called the complementary solutions, describing the general solution to the homogeneous version of the original differential equation. Complementary solutions are significant because they form the basis for constructing the solution to the non-homogeneous differential equation using methods like variation of parameters.

Wronskian

The Wronskian is a determinant used to assess whether a set of functions are linearly independent. For the complementary solutions \(y_1 = x + 2\) and \(y_2 = \frac{1}{x + 1}\), the Wronskian \(W(y_1, y_2)\) is calculated. It's an essential step because if the Wronskian is not zero, it confirms the linear independence of the solutions, making them valid for constructing a particular solution. In this exercise, the Wronskian is calculated as: \[ W(y_1, y_2) = \begin{vmatrix} x+2 & \frac{1}{x+1} \ 1 & \frac{-1}{(x+1)^2} \end{vmatrix} = -\frac{1}{(x+1)^3} - \frac{(x + 2)}{(x+1)^2} \] The result is crucial for proceeding with the variation of parameters.

Particular Solution

The particular solution is a specific solution that fits the non-homogeneous part of the differential equation. Unlike the complementary solution, which handles the homogeneous part, the particular solution addresses the entire equation. Using the method of variation of parameters, we determine functions \(u_1\) and \(u_2\) that, when combined with complementary solutions, result in the particular solution. The steps involve calculating \(u_1'\) and \(u_2'\) by solving a system of linear equations derived from substituting into the original non-homogeneous equation.

Once these values are determined, they are integrated to obtain \(u_1\) and \(u_2\). As seen in the solution, this results in:

• \(u_1(x) = -\frac{1}{(x+1)^2} + c_1 \)
• \(u_2(x) = -\frac{x^{2}}{4(x+1)^{2}} + \frac{x}{2(x+1)} - 1 + c_2\)

These functions are then used to construct the full solution as \( y = y_p = u_1y_1 + u_2y_2 \), tailored to the initial and boundary conditions.