Question

In Exercises \(24-29\) use the principle of superposition to find a particular solution. $$ y^{\prime \prime}+y=e^{-x}\left(2-4 x+2 x^{2}\right)+e^{3 x}\left(8-12 x-10 x^{2}\right) $$

Short answer

What is the particular solution to the differential equation: $$ y^{\prime \prime}+y=e^{-x}\left(2-4 x+2 x^{2}\right)+e^{3 x}\left(8-12 x-10x^{2}\right) $$ Answer: The particular solution to the given differential equation is: $$ y_p(x) = e^{-x}(2x^2 + 4x + 6) + e^{3x}(-10x^2 - 18x + 14) $$

Step-by-step solution

Identify the forcing functions

For this given differential equation, we have two separate forcing functions: 1. $$ F_1(x) = e^{-x}(2-4x+2x^2) $$ 2. $$ F_2(x) = e^{3x}(8-12x-10x^2) $$

Compute the particular solution for each forcing function

We will now find the particular solution for each forcing function separately: (2.1) $$ y_1''(x) + y_1(x) = e^{-x}(2-4x+2x^2) $$ Since we have the quadratic term within the exponential term, we should try the ansatz for the particular solution to be the form of: $$ y_{1p}(x) = e^{-x}(ax^2 + bx + c) $$ (2.2) $$ y_2''(x) + y_2(x) = e^{3x}(8-12x-10x^2) $$ Similarly, for the second forcing function, the ansatz to be tried for the particular solution is: $$ y_{2p}(x) = e^{3x}(dx^2 + ex + f) $$

Use the ansatz and calculate derivatives

Differentiate the ansatz solutions for each equation and substitute them into the corresponding equation: (3.1) Compute the derivatives for $$ y_{1p}(x) = e^{-x}(ax^2 + bx + c) $$: $$ y_{1p}'(x) = e^{-x}[ax^2 + (2a-b)x +(b-c-1)] $$ $$ y_{1p}''(x) = e^{-x}[ax^2 + (4a-2b)x +(4a-b-c+2)] $$ Then, substitute it into Equation (2.1): $$ e^{-x}[ax^2 + (4a-2b)x +(4a-b-c+2)] + e^{-x}(ax^2 + bx + c) = e^{-x}(2-4x+2x^2)$$ (3.2) Compute the derivatives for $$ y_{2p}(x) = e^{3x}(dx^2 + ex + f) $$: $$ y_{2p}'(x) = e^{3x}[dx^2 + (2d-e)x +(3d+e-3)] $$ $$ y_{2p}''(x) = e^{3x}[dx^2 + (6d-2e)x +(9d-3e-6)] $$ Then, substitute it into Equation (2.2): $$ e^{3x}[dx^2 + (6d-2e)x +(9d-3e-6)] + e^{3x}(dx^2 + ex + f) = e^{3x}(8-12x-10x^2) $$

Solve for the coefficients

Equating the coefficients for each equation: (4.1) For Equation (3.1): $$ a = 2 $$ $$ 4a - 2b = -4 \Rightarrow b = 4 $$ $$ 4a - b - c + 2 = 0 \Rightarrow c = 6 $$ (4.2) For Equation (3.2): $$ d = -10 $$ $$ 6d - 2e = -12 \Rightarrow e = -18 $$ $$ 9d - 3e - 6 = 8 \Rightarrow f = 14 $$

Combine the particular solutions

Now, we can write the particular solutions for each part: $$ y_{1p}(x) = e^{-x}(2x^2 + 4x + 6) $$ $$ y_{2p}(x) = e^{3x}(-10x^2 - 18x + 14) $$ Then, we can use the principle of superposition to combine them and obtain the particular solution to the original equation: $$ y_{p}(x) = y_{1p}(x) + y_{2p}(x) = e^{-x}(2x^2 + 4x + 6) + e^{3x}(-10x^2 - 18x + 14) $$ Finally, the particular solution to the given differential equation is: $$ y_p(x) = e^{-x}(2x^2 + 4x + 6) + e^{3x}(-10x^2 - 18x + 14) $$

Key concepts

Particular Solution

When dealing with differential equations, a particular solution is a solution that specifically satisfies a non-homogeneous differential equation, representing the particular case under given conditions. Unlike the general solution which includes arbitrary constants and represents an entire family of solutions, the particular solution resolves the equation for the fixed functions and derivatives involved.

Finding a particular solution often begins by identifying the parts of the differential equation that make it non-homogeneous, usually referred to as the "forcing function". By effectively handling these parts, using a systematic approach such as the superposition principle, we are able to find particular solutions tailored to these specific functions.

Forcing Function

The term "forcing function" in a differential equation represents the non-homogeneous part of the equation or the external input that causes changes or drives the system. The given problem involves two forcing functions:

• \( F_1(x) = e^{-x}(2-4x+2x^2) \)
• \( F_2(x) = e^{3x}(8-12x-10x^2) \)

These functions incorporate exponential and polynomial terms which directly influence the differential equation, making it necessary to separately address each forcing function to find its respective particular solution.

In practice, identifying and properly managing the forcing function is crucial as it dictates how the particular solution transforms the system described by the equation.

Differential Equation

A differential equation is a mathematical equation that involves unknown functions and their derivatives, depicting how these functions change with respect to variables. In the provided exercise, the equation is:\[ y'' + y = e^{-x}(2-4x+2x^2) + e^{3x}(8-12x-10x^2)\]This second-order differential equation with constant coefficients includes two different forcing functions on the right-hand side. Handling this equation requires addressing both stability and behavior of solutions under external influences.

By addressing differential equations, we model real-world phenomena such as motion, heat distribution, or population dynamics, allowing us to predict behaviors of complex systems.

Ansatz Method

The ansatz method is a powerful technique used to make educated guesses or propose a form for the particular solution of a differential equation. In this exercise, it was applied to each part corresponding to a forcing function:

• For \( F_1(x) = e^{-x}(2-4x+2x^2) \), the ansatz is \( y_{1p}(x) = e^{-x}(ax^2 + bx + c) \)
• For \( F_2(x) = e^{3x}(8-12x-10x^2) \), the ansatz is \( y_{2p}(x) = e^{3x}(dx^2 + ex + f) \)

Through differentiation and substitution, the coefficients are determined to fit the given equation. This method relies heavily on recognizing patterns and is crucial for simplifying complex differential equations into solvable problems.

By applying the ansatz, we effectively tailor the particular solution to align with the unique structure of the forcing functions presented.