Question

Use variation of parameters to solve the initial value problem, given \(y_{1}, y_{2}\) are solutions of the complementary equation. $$ \begin{array}{l} (x-1)^{2} y^{\prime \prime}-\left(x^{2}-1\right) y^{\prime}+(x+1) y=(x-1)^{3} e^{x}, \quad y(0)=4, \quad y^{\prime}(0)=-6 \\ y_{1}=(x-1) e^{x}, \quad y_{2}=x-1 \end{array} $$

Short answer

Based on the step-by-step solution above, answer the following question: **Question:** Find the general solution to the initial value problem \(y'' - 2y'= (x-1)^3 e^x\), with initial conditions \(y(0) = 4\) and \(y'(0) = -6\). **Answer:** The general solution to the initial value problem is given by: $$ y(x) = -5(x-1)e^x + (x-1) + y_p(x), $$ where \(y_p(x)=u(x-1)e^x + v(x-1),\) and \(u\) and \(v\) are given by $$ u = -\int \frac{(x-1)^3 e^x (x-1)}{(x-1)e^x - (x-1)(xe^x)}dx,\quad v = \int \frac{(x-1)^3 e^x (x-1)e^x}{(x-1)e^x - (x-1)(xe^x)}dx. $$

Step-by-step solution

Calculate the Wronskian of \(y_1\) and \(y_2\)

Compute the Wronskian \(W(x)\) as follows: $$ W(x) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} $$ where \(y_1 = (x-1)e^x\) and \(y_2 = x-1\). First, compute the derivatives: $$ y_1' = e^x(x-1) + e^x = e^x(x),\quad y_2' = 1 $$ Now, compute the determinant and thus obtain the Wronskian: $$ W(x) = \begin{vmatrix} (x-1)e^x & x-1 \\ xe^x & 1 \end{vmatrix} = (x-1)e^x - (x-1)(xe^x) $$

Find the particular integral \(y_p\)

The particular integral is given by the formula: $$ y_p = -y_1\int \frac{g(x) y_2}{W(x)}dx + y_2\int \frac{g(x) y_1}{W(x)}dx $$ Substitute \(g(x) = (x-1)^3 e^x\), \(y_1 = (x-1)e^x\), \(y_2 = x-1\), and \(W(x) = (x-1)e^x - (x-1)(xe^x)\) from steps 1: $$ y_p = -(x-1)e^x\int \frac{(x-1)^3 e^x (x-1)}{(x-1)e^x - (x-1)(xe^x)}dx + (x-1)\int \frac{(x-1)^3 e^x (x-1)e^x}{(x-1)e^x - (x-1)(xe^x)}dx $$ Integrate the two parts, we have: $$ y_p = u(x-1)e^x + v(x-1), $$ where $$ u = -\int \frac{(x-1)^3 e^x (x-1)}{(x-1)e^x - (x-1)(xe^x)}dx,\quad v = \int \frac{(x-1)^3 e^x (x-1)e^x}{(x-1)e^x - (x-1)(xe^x)}dx. $$

Apply the boundary conditions

Now, let's apply the initial conditions \(y(0)=4\) and \(y'(0)=-6\). First, the general solution will be $$ y(x) = c_1 (x-1)e^x + c_2 (x-1) +y_p(x). $$ Now, we apply the initial condition \(y(0)=4\) $$ 4=c_1(-1)e^0+c_2(-1)+0 $$ and \(y'(0) =-6\): $$ -6 = c_1 (e^0) + c_2 (1) + (0). $$

Solve for the constants

Solve the system of linear equations to find the constants \(c_1\) and \(c_2\): $$ \begin{cases} 4 = -c_1 - c_2 \\ -6 = c_1+c_2 \end{cases} $$ Solving the system, we obtain \(c_1 = -5\), and \(c_2 = 1\).

Write the solution

Finally, write down the solution: $$ y(x) = -5(x-1)e^x + (x-1) +y_p(x), $$ where \(y_p(x)=u(x-1)e^x + v(x-1),\) and \(u\) and \(v\) are given by $$ u = -\int \frac{(x-1)^3 e^x (x-1)}{(x-1)e^x - (x-1)(xe^x)}dx,\quad v = \int \frac{(x-1)^3 e^x (x-1)e^x}{(x-1)e^x - (x-1)(xe^x)}dx. $$

Key concepts

Wronskian

The Wronskian is a crucial concept when solving differential equations using variation of parameters. It helps determine whether a set of functions are linearly independent, which is vital for constructing a particular solution. In this context, given two functions, \(y_1\) and \(y_2\), the Wronskian is the determinant of a matrix constructed from these functions and their derivatives.

To compute the Wronskian for the functions \(y_1 = (x-1)e^x\) and \(y_2 = x-1\), you form a 2x2 matrix using these functions and their derivatives:

• Calculate \(y_1'\): This is the derivative of \((x-1)e^x\), using product rule it becomes \(e^x(x)\).
• Calculate \(y_2'\): The derivative of \(x-1\) is simply 1.

Next, arrange them in the matrix:\[W(x) = \begin{vmatrix} (x-1)e^x & x-1 \ xe^x & 1 \end{vmatrix} = (x-1)e^x \cdot 1 - (x-1)(xe^x)\]The simplification of this determinant will lead you to find that the Wronskian is not zero, confirming that \(y_1\) and \(y_2\) are indeed linearly independent. This is a necessary step to ensure that a particular solution can be found using these functions.

Particular Integral

The particular integral is a specific solution used in the method of variation of parameters. It solves the non-homogeneous part of a differential equation. The whole purpose is to determine a function that satisfies the original differential equation, but for the non-homogeneity present in it.

Here, the particular integral, \(y_p\), uses the previously calculated Wronskian along with other functions. It can be expressed in terms of the given functions \(y_1\) and \(y_2\):\[y_p = -y_1\int \frac{g(x) y_2}{W(x)}dx + y_2\int \frac{g(x) y_1}{W(x)}dx\]For the problem, substitute:

• \(g(x) = (x-1)^3 e^x\): This function represents the non-homogeneous part.
• \(W(x)\), \(y_1\), and \(y_2\) were calculated in the earlier steps.

Performing these integrations will yield functions \(u\) and \(v\), which when multiplied by \(x-1)e^x\) and \(x-1\) respectively, give us the particular integral. This helps in shaping the full solution to the differential equation by adding it to the complementary solution.

Initial Value Problem

The initial value problem (IVP) is a type of differential equation problem where the solution must satisfy specified conditions at the start. It is essential for obtaining a solution that not only satisfies the differential equation but also meets certain starting criteria.

In our problem, the IVP specified the conditions \(y(0)=4\) and \(y'(0)=-6\). This involves solving for constants that need to be determined from the general solution of the differential equation.

• The general solution form is \(y(x) = c_1 (x-1)e^x + c_2 (x-1) + y_p(x)\), where \(y_p(x)\) is the particular integral.
• Use \(y(0)=4\): Substitute into the general solution to find a relationship between \(c_1\) and \(c_2\).
• Use \(y'(0) = -6\) as another equation to solve the system of linear equations for \(c_1\) and \(c_2\).

By solving the system of equations, we ensure that the solution specifically passes through these initial conditions. It's an essential part of confirming that the solution meets all the problem's requirements and constraints.