Question

Refer to the cited exercises and use the principal of superposition to find a particular solution. Then find the general solution. $$ y^{\prime \prime}-4 y^{\prime}+4 y=2+8 x-4 x^{2}+e^{x} $$

Short answer

The general solution of the given non-homogeneous ordinary differential equation is: $$ y(x) = C_1 e^{2x} + C_2 xe^{2x} + 2 + 2x - 2x^2 + x e^{x} $$ Where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Convert the ODE into a homogeneous form

We are given the ODE in the form: $$ y^{\prime \prime}-4 y^{\prime}+4 y=2+8 x-4 x^{2}+e^{x} $$ We can notice that the given ODE is already in a homogeneous form. The left-hand side contains the terms with derivatives and y, while the right-hand side contains the terms 2, 8x, -4x^2, and e^x. So, no conversion is needed.

Identify the roots of the characteristic equation

For the homogeneous part of the ODE, we write the characteristic equation: $$ r^2 - 4r + 4 = 0 $$ Factoring this quadratic equation gives: $$ (r - 2)^2 = 0 $$ The roots of the characteristic equation are r1 = r2 = 2, which means the characteristic equation has a single, repeated root.

Solve the homogeneous equation and express the general solution

Since we have a single, repeated root for the characteristic equation (r1 = r2 = 2), the general solution of the homogeneous equation is given by: $$ y_h(x) = C_1 e^{2x} + C_2 xe^{2x} $$ Where \(C_1\) and \(C_2\) are constants.

Construct a particular solution using the non-homogeneous term

Using the principle of superposition, we can construct the particular solution by guessing a solution form for each term in the right-hand side (2, 8x, -4x^2, and e^x). Here, we will guess the particular solutions as follows: For the 2 term: \(A\), with A constant For the 8x term: \(Bx\), with B constant For the -4x^2 term: \(Cx^2\), with C constant For the e^x term: \(Dxe^{x}\), with D constant (since e^x is already included in the homogeneous solution) Combining these guesses, the ansatz for the particular solution is: $$ y_p(x) = A + Bx + Cx^2 + Dxe^{x} $$ Now, we need to differentiate y_p(x) twice to substitute it in the ODE and find the constants A, B, C, and D. $$ y_p^{\prime}(x) = B + 2Cx + De^{x} + Dxe^{x} $$ $$ y_p^{\prime\prime}(x) = 2C + De^{x} + De^{x} + Dxe^{x} $$ Now substitute \(y_p\), \(y_p^{\prime}\) and \(y_p^{\prime\prime}\) in the given ODE and compare the coefficients to obtain a system of linear equations for the constants A, B, C, and D. Solve these equations to find the values of A, B, and C. The non-homogeneous ODE becomes: $$ (2C + De^{x} + De^{x} + Dxe^{x}) - 4(B + 2Cx + De^{x} + Dxe^{x}) + 4(A + Bx + Cx^2 + Dxe^{x}) = 2 + 8x - 4x^2 + e^x $$ Equating the coefficients, we obtain the following system of linear equations: $$ A - 4B + 4C = 2 $$ $$ B - 8C + 4D = 8 $$ $$ De^x = e^x $$ $$ 2C = -4 $$ Solving these equations, we find A = 2, B = 2, C = -2, and D = 1. Thus, the particular solution is: $$ y_p(x) = 2 + 2x - 2x^2 + x e^{x} $$

Sum the homogeneous general solution with the particular solution

Finally, we sum the general solution of the homogeneous equation and the particular solution to obtain the general solution of the given non-homogeneous ODE: $$ y(x) = y_h(x) + y_p(x) = C_1 e^{2x} + C_2 xe^{2x} + 2 + 2x - 2x^2 + x e^{x} $$ This is the general solution of the given non-homogeneous ODE, where \(C_1\) and \(C_2\) are arbitrary constants.

Key concepts

Superposition Principle

The Superposition Principle is a fundamental concept in solving linear differential equations, especially when dealing with non-homogeneous cases. This principle states that the solution to a linear differential equation can be represented as the sum of the solution to its homogeneous part and a particular solution corresponding to the non-homogeneous part.
This is because any linear combination of solutions to a linear differential equation is also a solution.
When given a non-homogeneous differential equation like \( y'' - 4y' + 4y = 2 + 8x - 4x^2 + e^x \), we start by solving for the homogeneous part separately. Then, using the principle of superposition, we find a particular solution for the non-homogeneous part. The general solution is then the sum of these two solutions.
This approach simplifies the process as it breaks down the problem into manageable parts:

• Solving the homogeneous equation
• Finding a particular solution
• Adding these solutions together

This method makes it easier to handle complex differential equations by leveraging the linearity property.

Characteristic Equation

The Characteristic Equation is critical when you want to solve a homogeneous linear differential equation. For our problem, we derive this equation from the homogeneous part \( y'' - 4y' + 4y = 0 \).
This involves a substitution where we assume solutions of the form \( y = e^{rx} \), leading to a polynomial equation in terms of \( r \). Solving this equation gives us the roots that dictate the form of the homogeneous solution.
For example, the characteristic equation here is \( r^2 - 4r + 4 = 0 \). By factoring it, we get \((r - 2)^2 = 0\), which means it has a repeated root \( r = 2 \).
When roots are repeated, the general solution for the homogeneous equation does not only include terms like \( C_1 e^{rx} \), but also \( C_2 xe^{rx} \) to account for the multiplicity of the root.
Thus, in our solution, the homogeneous part is \( y_h(x) = C_1 e^{2x} + C_2 xe^{2x} \), where \( C_1 \) and \( C_2 \) are constants determined by initial or boundary conditions if provided.

Non-Homogeneous Differential Equation

A Non-Homogeneous Differential Equation includes terms that are not solely dependent on the function and its derivatives. This means the equation has an external force or term, such as \( 2 + 8x - 4x^2 + e^x \), making the differential equation non-homogeneous.
This addition on the right side compels us to find a particular solution distinct from the homogeneous solution. It requires evaluating the form of each non-homogeneous term.
In practice, a common technique is the method of undetermined coefficients or variation of parameters. In this example, we use undetermined coefficients by trial functions like \( A \) (constant), \( Bx \), \( Cx^2 \), and \( Dxe^x \).
These trial functions correspond to each term in the non-homogeneous part:

• \(2\) corresponds to \(A\)
• \(8x\) corresponds to \(Bx\)
• \(-4x^2\) corresponds to \(Cx^2\)
• \(e^x\) corresponds to a modified term \(Dxe^x\) since \(e^x\) appears in the homogeneous solution.

This enables constructing a suitable particular solution \( y_p(x) = 2 + 2x - 2x^2 + xe^x \), ensuring continuity and differentiation match the original equation.

Particular Solution

The Particular Solution is a vital part of solving a non-homogeneous differential equation. It effectively targets the non-homogeneous part, crafting a solution that inherently matches the extra terms on the right side.
In this exercise, we have the non-homogeneous portion \( 2 + 8x - 4x^2 + e^x \). To find the particular solution, we hypothesize a solution form based on these terms. We then determine unknown coefficients by plugging this form into the main equation and aligning coefficients.
For instance, with the guess \(y_p(x) = A + Bx + Cx^2 + Dxe^x\):
- For constant \(2\), we set \(A\)- For linear \(8x\), we use \(Bx\)- For quadratic \(-4x^2\), we apply \(Cx^2\)- The exponential term \(e^x\) uses \(Dxe^x\) due to overlap with the homogeneous solution
Once derived, this specific solution is combined with the homogeneous solution to form the complete general solution, which in this exercise is \(y(x) = C_1 e^{2x} + C_2 xe^{2x} + 2 + 2x - 2x^2 + xe^x\).
This general solution encompasses all potential solutions to the differential equation, addressing both homogeneous and non-homogeneous components.