Question

Use the principle of superposition to find a particular solution. Where indicated, solve the initial value problem. $$ y^{\prime \prime}+y=4 \cos x-2 \sin x+x e^{x}+e^{-x} $$

Short answer

Answer: The principle of superposition states that if we have two independent solutions to a linear differential equation, their linear combination is also a solution. To find a particular solution using the principle of superposition, we first find individual particular solutions corresponding to each of the functions on the right-hand side of the equation. Then, we add these individual solutions together to obtain the complete particular solution for the given differential equation.

Step-by-step solution

Find the derivatives of the given equation

First, find the derivatives of the given equation, which is: $$ y^{\prime \prime}+y=4 \cos x-2 \sin x+x e^{x}+e^{-x} $$

Find the particular solution for \(4\cos{x}\)

Let's find the particular solution for \(4\cos{x}\). Let \(y_p=A\cos{x} + B\sin{x}\). Find the first and second derivatives of \(y_p\). Then, substitute these results into the given equation and solve for \(A\) and \(B\). The resulting expression will be the particular solution corresponding to \(4\cos{x}\).

Find the particular solution for \(-2\sin{x}\)

Now, let's find the particular solution for \(-2\sin{x}\). We can use the result found in the previous step to create the particular solution for this case. Since we have already found \(A\) and \(B\), we can replace them with the necessary constants and construct the new particular solution.

Find the particular solution for \(xe^x\)

To find the particular solution for \(xe^x\), we will assume the form of the particular solution to be \(y_p = (Ax+B)e^x\). Now, find the first and second derivatives of this assumed form and substitute these into the given differential equation. Solve for \(A\) and \(B\) and replace them by the resulting constants.

Find the particular solution for \(e^{-x}\)

Finally, let's find the particular solution for \(e^{-x}\). For this case, we will assume the form of the particular solution to be \(y_p = (Cx+D)e^{-x}\). Derive the first and second derivatives of this equation and substitute them into the given differential equation. Solve for \(C\) and \(D\) and replace them by the resulting constants.

Combine all the particular solutions

Now that we have found the particular solutions for each term on the right-hand side (i.e., \(4\cos{x}\), \(-2\sin{x}\), \(xe^x\) and \(e^{-x}\)), all that remains is to add them together. This will give us the complete particular solution to the given differential equation. The combined particular solution will have the form: $$ y_p = (A\cos{x} + B\sin{x}) + (-2\sin{x}) + ((Ax+B)e^x) + ((Cx+D)e^{-x}) $$ where \(A\), \(B\), \(C\), and \(D\) are the constants that we found in the previous parts.

Key concepts

Particular Solution Differential Equations

In the study of differential equations, finding a particular solution involves determining a single function that satisfies the equation for specific non-homogeneous conditions—in this case, the presence of terms such as \(4 \cos x-2 \sin x+x e^{x}+e^{-x}\). The superposition principle is essential here; it allows us to find the particular solution for each individual term on the right side of the equation and combine them into a complete solution. This principle operates under the assumption that the differential equation is linear.

When each unique input is considered—which might be functions like trigonometric functions or exponentials—the associated particular solution is structured to match the form of the input. For example, with the term \(4\cos x\), a corresponding particular solution might take the form of \(A\cos x + B\sin x\) with constants \(A\) and \(B\) to be determined. The term \(-2\sin x\), being a sine function, also contributes a sinusoidal component to the particular solution.

In contrast, exponential term solutions will appear as adjustments of the factor in front of their respective exponential functions. To solve for these constants, we use methods like substitution and equate coefficients. This process is repeated for each term until all constants are found, and then, as per the superposition principle, these individual solutions are summed to create the overall particular solution.

Initial Value Problem

An initial value problem (IVP) is a specific type of differential equation which not only seeks a general solution but also requires the solution to satisfy initial conditions. These conditions are the values of the unknown function or its derivatives at a certain point. Looking at the original exercise, the instruction may include solving the initial value problem if with given initial conditions, indicating the necessity to adjust the general solution to satisfy those conditions.

Initial conditions play a critical role in determining the constants within the general solution of a differential equation. Without them, you may have a multitudinous set of solutions. Once initial conditions are applied, they pinpoint the accurate particular solution that fits the provided conditions—essentially allowing the differential equation to model real-world scenarios with precision.

For example, if you are given the IVP where \( y(0) \) or \( y'(0) \) is provided, these conditions would be used in conjunction with the general solution to create a unique equation that not only solves the differential equation but also passes through the specified points on the graph of the solution, anchoring the 'C' values in a unique particular solution.

Solving Differential Equations

Solving differential equations, particularly second-order non-homogeneous equations—as depicted in our exercise—requires a systematic approach. Key steps include finding a complementary solution that serves the homogeneous form of the equation and a particular solution that accounts for the non-homogeneous terms.

To achieve this, one may use methods such as undetermined coefficients, as seen in our step-by-step solution, where guesses of solutions are made based on the form of the non-homogeneous term. Another approach is the method of variation of parameters. Once appropriate forms for the particular solution are guessed, their derivatives are computed, and these forms are substituted back into the differential equation to solve for unknowns.

Ultimately, the solution to a given differential equation is typically the sum of the complementary and particular solutions. The real work lies in carefully crafting the particular solution to reflect the input functions on the non-homogeneous side of the equation and subsequently applying any initial conditions to resolve constants for a specific solution—core steps that tie the entire process together, guiding students from problem state to solution.