Question

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ (x-1) y^{\prime \prime}-x y^{\prime}+y=2(x-1)^{2} e^{x} ; \quad y_{1}=x, \quad y_{2}=e^{x} $$

Short answer

Question: Find a particular solution to the non-homogeneous second-order differential equation \(x^2y'' - xy' + y = 2(x - 1)^2e^x\) using the method of variation of parameters. The solutions of the complementary equation are \(y_1 = x\) and \(y_2 = e^x\). Answer: The particular solution to the given non-homogeneous second-order differential equation is \(y_p(x) = -2x^2 - 3xe^x - e^{2x} + C_1e^x + C_2x\), where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Calculate the Wronskian of the complementary solutions

Given the complementary solutions, \(y_1 = x\) and \(y_2 = e^x\). Let's calculate their Wronskian, \(W(y_1, y_2)\). The Wronskian is defined as: $$ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} $$ So, we first need to find the derivatives of \(y_1\) and \(y_2\): $$ y_1' = 1, \quad y_2' = e^x $$ Now, we can find the Wronskian: $$ W(y_1, y_2) = \begin{vmatrix} x & e^x \\ 1 & e^x \end{vmatrix} = xe^x - e^x = e^x(x - 1) $$

Apply the variation of parameters formula

According to the variation of parameters method, we assume that the particular solution is of the form: $$ y_p(x) = u_1(x)y_1 + u_2(x)y_2 $$ We need to find the functions \(u_1(x)\) and \(u_2(x)\). To do this, we first need the first derivatives of \(u_1\) and \(u_2\): $$ u_1'(x)y_1 + u_2'(x)y_2 = -\frac{f(x)}{W} $$ Here, \(f(x) = 2(x - 1)^2e^x\). Therefore, $$ u_1'(x)x + u_2'(x)e^x = -\frac{2(x - 1)^2e^x}{e^x(x - 1)} = -2(x - 1) $$ We obtain a system of equations: $$ \begin{cases} u_1'(x)x = -2(x - 1) + u_2'(x)e^x \\ u_2'(x)e^x = (2(x - 1) - u_1'(x)x) \end{cases} $$

Solve for \(u_1'(x)\) and \(u_2'(x)\) and integrate

Now, we can integrate both equations to find \(u_1(x)\) and \(u_2(x)\): $$ \begin{cases} u_1(x) = \int{-2 + \frac{u_2'(x)e^x}{x}} \,dx \\ u_2(x) = \int{\frac{2(x - 1)}{e^x} - \frac{u_1'(x)x}{e^x}} \,dx \end{cases} $$ To solve the integrals, we can use integration by parts. Let's solve the second integral first: $$ u_2(x) = \int{\frac{2(x - 1)}{e^x}} \,dx - \int{\frac{u_1'(x)x}{e^x}} \,dx $$ Taking the first term, we have $$ u_2(x) = -2e^{-x}(x - 1) - e^{-x} + C_1 $$ Now, solve the first integral for \(u_1(x)\): $$ u_1(x) = \int{-2 + \frac{2e^x(x - 2) - u_1'(x)e^x}{x}} \,dx = -2x - 2e^x + C_2 $$

Determine the particular solution

Now that we have found \(u_1(x)\) and \(u_2(x)\), we can substitute them into the particular solution: $$ y_p(x) = u_1(x)y_1 + u_2(x)y_2 = (-2x - 2e^x + C_2)(x) + (-2e^{-x}(x - 1) - e^{-x} + C_1)(e^x) $$ Simplifying gives the final particular solution: $$ y_p(x) = -2x^2 - 3xe^x - e^{2x} + C_1e^x + C_2x $$

Key concepts

Wronskian

The Wronskian is a crucial determinant when analyzing differential equations, as it helps to verify if the solution set is linearly independent. Let's consider two functions, say, the complementary solutions, \( y_1 = x \) and \( y_2 = e^x \). The Wronskian for these functions is calculated using their derivatives.
It is defined as:

• The matrix: \( W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} \)
• For our functions: \( y_1' = 1 \) and \( y_2' = e^x \).

Substitute these into the Wronskian to get \( W = xe^x - e^x \). Simplifying results in \( e^x(x-1) \).
A non-zero Wronskian confirms that the functions \( y_1 \) and \( y_2 \) are linearly independent solutions of the differential equation, which is vital for advanced calculations like Variation of Parameters.

Complementary Solutions

Complementary solutions are solutions to the homogeneous part of a differential equation. For our problem, the differential equation is given as \( (x-1)y'' - xy' + y = 0 \). These solutions form the backbone of our approach when solving the non-homogeneous equation.

The complementary solution is typically expressed as a linear combination of the two independent solutions. For our equation, the complementary solutions are \( y_1 = x \) and \( y_2 = e^x \). Thus, any solution to the homogeneous equation can be written as:

• \( y_c = C_1 y_1 + C_2 y_2 = C_1 x + C_2 e^x \)

Here, \( C_1 \) and \( C_2 \) are constants, dependent on initial conditions.
Utilizing these clearly defined solutions is critical since they set the stage for finding particular solutions of the non-homogeneous version.

Particular Solution

A particular solution refers to a specific solution of the non-homogeneous differential equation. By finding it, we ensure that our overall solution includes the effects of any external or non-homogeneous forcing function.
In variation of parameters, we assume solutions of the format:\[ y_p(x) = u_1(x)y_1 + u_2(x)y_2 \]
Here, \( u_1(x) \) and \( u_2(x) \) are functions to be determined. They add flexibility, going beyond what the complementary solutions offer.
Finding \( u_1 \) and \( u_2 \) involved the equations:

• \( u_1'(x)y_1 + u_2'(x)y_2 = -\frac{f(x)}{W} \)

For our problem, \( f(x) = 2(x-1)^2e^x \), and the Wronskian is \( W(y_1, y_2) = e^x(x-1) \). Solving these results in \( u_1 \) and \( u_2 \) which are thoroughly integrated and substituted back into the assumed format for the particular solution. Ultimately, we arrive at a particular expression: \( y_p(x) = -2x^2 - 3xe^x - e^{2x} + C_1e^x + C_2x \).
Having this detailed solution allows full coverage of forces acting on the system.

Integration by Parts

Integration by parts is a powerful integration technique used to integrate products of functions. It’s especially handy in situations involving combinations, like \( \int uv' \,dx \), and can be formalized as:

• \( \int u \, dv = uv - \int v \, du \)

In the method of variation of parameters, integration by parts is often necessary to solve for unknown functions \( u_1(x) \) and \( u_2(x) \).

When applied, you pick one part of the integrand as \( u \) and the rest as \( dv \). The choice of \( u \) often simplifies \( du \) and makes \( \int v \, du \) easier to handle. This was done to solve for \( u_2(x)\) in the particular solution derivation, where sometimes neither term is easily integrable on its own.
Using integration by parts unlocks solutions that seem trapped in complexity, highlighting why it is one of the preferred methods for handling certain challenging integrals.