Question

Find a particular solution, given the fundamental set of solutions of the complementary equation. $$ x^{4} y^{(4)}-4 x^{3} y^{\prime \prime \prime}+12 x^{2} y^{\prime \prime}-24 x y^{\prime}+24 y=x^{4} ; \quad\left\\{x, x^{2}, x^{3}, x^{4}\right\\} $$

Short answer

Answer: The particular solution to the given differential equation is \(y_p = \frac{1}{24}x^4\).

Step-by-step solution

Make a trial solution assumption

As the nonhomogeneous term is \(x^4\), we may guess the particular solution to be in the form: $$ y_p = Ax^4 $$ where A is a constant to be determined.

Calculate derivatives of the trial solution

We need to find the first, second, third, and fourth derivatives of our trial solution: First derivative: $$ y_p' = 4Ax^3 $$ Second derivative: $$ y_p'' = 12Ax^2 $$ Third derivative: $$ y_p''' = 24Ax $$ Fourth derivative: $$ y_p^{(4)} = 24A $$

Plug the trial solution and its derivatives into the given differential equation

Now we substitute our trial solution and its derivatives back into the given differential equation, $$ x^4 y^{(4)} - 4x^3 y''' + 12x^2 y'' - 24x y' + 24 y = x^4 $$ Substituting the derivatives, we have: $$ x^4(24A) - 4x^3(24Ax) + 12x^2(12Ax^2) - 24x(4Ax^3) + 24(Ax^4) = x^4 $$ Simplifying, $$ 24Ax^4 - 96Ax^4 + 144Ax^4 - 96Ax^4 + 24Ax^4 = x^4 $$

Solve for the unknown constant A

Combining terms, we get: $$ 24Ax^4 = x^4 $$ Divide both sides by \(x^4\): $$ 24A = 1 $$ Solve for A: $$ A = \frac{1}{24} $$

Write the particular solution

We now have the value of A, so we can write the particular solution: $$ y_p = \frac{1}{24}x^4 $$ Thus, the particular solution to the given differential equation is: $$ y_p = \frac{1}{24}x^4 $$

Key concepts

Particular Solution

A particular solution to a differential equation is a specific instance of solutions that satisfies the given equation with the nonhomogeneous part. Unlike the general solution, which encompasses a family of possible solutions, the particular solution is one unique solution that aligns with the special conditions or external forces described in the equation. In our example, we started by making an educated guess for the particular solution form. Given that the nonhomogeneous term was \( x^4 \), we assumed a trial solution:

• \( y_p = Ax^4 \)

Here we introduce "\( A \)" as an unknown constant that helps us find the exact solution. Then, by deriving this trial solution and substituting back into the differential equation, we solved for "\( A \)". This gave us our particular solution:

• \( y_p = \frac{1}{24}x^4 \)

The particular solution thus captures the specific behavior or influence described by the nonhomogeneous element \( x^4 \) in the differential equation.

Fundamental Set of Solutions

In the context of linear differential equations, a fundamental set of solutions refers to a set of linearly independent solutions that form the basis for the solution space of the corresponding homogeneous equation. This means that any solution to the homogeneous differential equation can be represented as a linear combination of these fundamental solutions. By using these solutions, we can solve for the full general solution of the differential equation.

For the example given:

• \( \{x, x^2, x^3, x^4\} \)

represents the fundamental set of solutions for our complementary equation. Each of these solutions is derived from the complementary equation (the homogeneous part) of the differential equation. This set is crucial for constructing the complementary solution \( y_c \), which will effectively address the homogeneous component. By combining the complementary and particular solutions, one obtains the general solution of the nonhomogeneous equation.

Complementary Equation

In solving differential equations, the complementary equation is derived from setting the nonhomogeneous part (external force or input) to zero. It helps ascertain the behavior of the system in the absence of external influences. The complementary solution \( y_c \) is then a linear combination of the fundamental set of solutions.

The complementary equation corresponding to our exercise would be written as:

• \( x^4 y^{(4)} - 4 x^3 y^{\prime \prime \prime} + 12 x^2 y^{\prime \prime} - 24 x y' + 24 y = 0 \)

This equation essentially captures the intrinsic dynamics of the system described by the differential equation without considering any external forcing term.

The general solution of the differential equation involves a sum of the complementary solution \( y_c \) and the particular solution \( y_p \). Therefore, solving for the complementary solution is a critical step when dealing with nonhomogeneous differential equations.

Nonhomogeneous Differential Equations

Nonhomogeneous differential equations differ from their homogeneous counterparts by including an external term or force. This external term introduces new characteristics to the solution that differ from the natural behavior of the system.

The differential equation in our situation,

• \( x^4 y^{(4)} - 4 x^3 y^{\prime \prime \prime} + 12 x^2 y^{\prime \prime} - 24 x y' + 24 y = x^4 \)

is nonhomogeneous because of the presence of \( x^4 \) on the right-hand side.The approach to solving such equations typically involves two primary steps:

• Finding the complementary solution by solving the associated homogeneous equation.
• Determining a particular solution that addresses the nonhomogeneous component.

By combining these solutions, you obtain the complete solution to the differential equation, which includes both the effects of the natural system behavior and the influences from external sources or conditions.