Problem 35 - Differential Equations and Linear Algebra

Question

Find the general solution to the given differential equation and the maximum interval on which the solution is valid. $$y^{\prime \prime}=x e^{x}$$.

Step-by-Step Solution

Answer

The general solution to the given differential equation \(y^{\prime \prime} = xe^x\) is \(y(x) = C_1 + C_2x + \frac{1}{8}x^3e^x\), where \(C_1\) and \(C_2\) are constants. The maximum interval on which the solution is valid is (-∞,∞).

Step 1: Assume a particular solution

We will assume a particular solution of the form \(y_p(x) = Ax^3e^x + Bx^2e^x\).

Step 2: Calculate the first and second derivatives

We will calculate the first and second derivatives of the assumed particular solution. The first derivative, \(y_p'(x)\), is: \[y_p'(x) = (3Ax^2e^x + 2Bxe^x) + (Ax^3e^x + Bx^2e^x)\] The second derivative, \(y_p''(x)\), is: \[y_p''(x) = (6Axe^x + 2Be^x) + (6Axe^x + 4Bxe^x) + (2Axe^x + 2Be^x)\] which simplifies to: \[y_p''(x) = (8A + 6B)xe^x + 2Be^x\]

Step 3: Substitute into the original equation and solve for coefficients

Now we substitute the second derivative into the original equation: \(y^{\prime \prime} = x e^x\) \((8A + 6B)xe^x + 2Be^x = xe^x\) Divide both sides by \(xe^x\): \(8A + 6B = 1\) Since there is no constant term on the right side of the equation, this implies: \(2B = 0 => B = 0\) Now we can solve for A using the first equation: \(8A + 6(0) = 1 => 8A = 1 => A = \frac{1}{8}\) Now we have determined that the particular solution \(y_p(x) = \frac{1}{8}x^3e^x\).

Step 4: Solve the homogeneous equation

Next, we need to solve the homogeneous equation \(y^{\prime \prime} = 0\). The general solution to this equation is \(y_c(x) = C_1 + C_2x\), where \(C_1\) and \(C_2\) are constants.

Step 5: Combine complementary and particular solutions

Finally, the general solution to the given differential equation is the sum of the complementary and particular solutions: \[y(x) = y_c(x) + y_p(x) = C_1 + C_2x + \frac{1}{8}x^3e^x\] Since our general solution does not have any singularities or undefined intervals, the maximum interval on which the solution is valid is (-∞,∞).

Key Concepts

General Solution

In differential equations, a general solution combines all possible solutions to a differential equation. It typically involves constants that can be adjusted based on initial conditions or other constraints. In the context of the differential equation provided, the general solution is formed by adding a particular solution and the complementary (or homogeneous) solution.
The particular solution satisfies the non-homogeneous part of the equation, which in this exercise is the part with the function on the right side of the equation, specifically, \(xe^x\). The complementary solution deals with the homogeneous equation, which has a zero on the right side.
Once both solutions are found, they are combined to create the general solution. Such a solution encompasses all specific solutions that could satisfy the original differential equation. In this case, the general solution is:

\[y(x) = C_1 + C_2x + \frac{1}{8}x^3e^x\]
where \(C_1\) and \(C_2\) are arbitrary constants reflecting the family of solutions.

Particular Solution

The particular solution of a differential equation addresses the non-homogeneous part of the equation. It is a specific solution that satisfies the differential equation for all values of the independent variable. To find a particular solution, assume a form that mirrors the non-homogeneous part. For example, if the non-homogeneous term is \(xe^x\), a reasonable assumption could be \(Ax^3e^x + Bx^2e^x\).
This assumed form is then differentiated as needed and substituted back into the original differential equation. By doing so, you solve for the coefficients in the assumed form, ensuring that the derived solution satisfies the equation.
For the current exercise, the assumed particular solution \(y_p(x) = \frac{1}{8}x^3e^x\) fits the form needed to address the non-homogeneous term of \(xe^x\). By using mathematical techniques such as taking derivatives and substituting, you ensure that \(y_p(x)\) specifically solves the differential equation.

Homogeneous Equation

A homogeneous equation in differential equations is one where all terms contain the dependent variable or its derivatives. No free-standing constants or independent variables appear on the right side. In our problem, the homogeneous part is represented by \(y'' = 0\), which indicates the second derivative may sum to zero.
The general solution to a homogeneous equation is composed of a linear combination of solutions, usually involving constants. Solving \(y'' = 0\) reveals the complementary solution \(y_c(x) = C_1 + C_2x\). This solution forms the backbone of the general solution due to its ability to account for initial conditions or boundary values by adjusting the constants \(C_1\) and \(C_2\).
Including the complementary solution ensures that once specific conditions are introduced, the final solution remains valid across the intended interval. This completes the composition of the general solution, ensuring that all aspects of the differential equation, homogeneous and otherwise, are addressed.