Question

By the method of undetermined coefficients determine a particular solution of each of the following equations: (a) \(y^{\prime \prime}+y=e^{x}+x^{2}\). (b) \(y^{\prime \prime}+2 y^{\prime}+y=2+\sin x\). (c) \(y^{\prime \prime}-4 y^{\prime}+4 y=e^{2 x}\).

Short answer

Answer: The particular solutions for each of the given differential equations are: (a) \(y_p(x) = e^x\) (b) \(y_p(x) = 2 + \sin x - \frac{1}{2}\cos x\) (c) \(y_p(x) = \frac{1}{4}xe^{2x}\).

Step-by-step solution

Determine the form of the particular solution

For the first equation, we have the right-hand side as \(e^x + x^2\). The general form of the particular solution should resemble the right-hand side with undetermined coefficients. Therefore, the particular solution should have the form: \(y_p(x) = Ae^x + Bx^2 + Cx + D\).

Differentiate the particular solution twice

To replace \(y^{\prime \prime}\) and \(y\) in the equation, we need to find the first and second derivatives of the particular solution: \(y_p'(x) = Ae^x + 2Bx + C\) \(y_p''(x) = Ae^x + 2B\)

Plug in undetermined coefficients and derivatives

Now, we plug in our undetermined coefficients and derivatives into the given equation: \(y_p''(x) + y_p(x) = (Ae^x + 2B) + (Ae^x + Bx^2 + Cx + D) = e^x + x^2\)

Compare coefficients and solve system of equations

Comparing coefficients on both sides, we get the following system of equations: \(A = 1\) \(2B = 0\) \(A + B = 0\) \(C = 0\) \(D = 0\) Solving this system, we find the undetermined coefficients to be \(A = 1\), \(B = 0\), \(C = 0\), and \(D = 0\).

Write down the particular solution

Using the values for undetermined coefficients, the particular solution is: \(y_p(x) = e^x\) This is the final particular solution for the first given equation. #(b) Equation: \(y^{\prime \prime} + 2y^{\prime} + y = 2 + \sin x\)

Determine the form of the particular solution

For the second equation, we have the right-hand side as \(2 + \sin x\). The general form of the particular solution should resemble the right-hand side with undetermined coefficients. Therefore, the particular solution should have the form: \(y_p(x) = A + B\sin x + C\cos x\).

Differentiate the particular solution twice

Calculate the first and second derivatives: \(y_p'(x) = B\cos x - C\sin x\) \(y_p''(x) = -B\sin x - C\cos x\)

Plug in undetermined coefficients and derivatives

Plug in the undetermined coefficients and derivatives: \(y_p''(x) + 2y_p'(x) + y_p(x) = (-B\sin x - C\cos x) + 2(B\cos x - C\sin x) + (A + B\sin x + C\cos x) = 2 + \sin x\)

Compare coefficients and solve system of equations

Comparing coefficients, we get: \(A = 2\) \(-B + 2C = 0\) \(B + 2C = 1\) Solving this system, we find the undetermined coefficients to be \(A=2\), \(B=1\), and \(C=-\frac{1}{2}\).

Write down the particular solution

Using the values for undetermined coefficients, the particular solution for the second equation is: \(y_p(x) = 2 + \sin x - \frac{1}{2}\cos x\) #(c) Equation: \(y^{\prime \prime} - 4y^{\prime} + 4y = e^{2x}\)

Determine the form of the particular solution

The right-hand side of the third equation is \(e^{2x}\), so the general form of the particular solution should be \(y_p(x) = Axe^{2x}\).

Differentiate the particular solution twice

Calculate the first and second derivatives: \(y_p'(x) = 2Axe^{2x} + Ae^{2x}\) \(y_p''(x) = 4Ax^2e^{2x} + 4Axe^{2x}\)

Plug in undetermined coefficients and derivatives

Plug in the undetermined coefficients and derivatives: \(y_p''(x) - 4y_p'(x) + 4y_p(x) = (4Ax^2e^{2x} + 4Axe^{2x}) - 4(2Axe^{2x} + Ae^{2x}) + 4(Axe^{2x}) = e^{2x}\)

Compare coefficients and solve system of equations

Comparing coefficients, we get: \(A = \frac{1}{4}\)

Write down the particular solution

Using the value for the undetermined coefficient, the particular solution for the third equation is: \(y_p(x) = \frac{1}{4}xe^{2x}\) These are the particular solutions for each of the given differential equations: (a) \(y_p(x) = e^x\) (b) \(y_p(x) = 2 + \sin x - \frac{1}{2}\cos x\) (c) \(y_p(x) = \frac{1}{4}xe^{2x}\)

Key concepts

Particular Solution

To solve a differential equation using the method of Undetermined Coefficients, we aim to find a particular solution. This is different from solving for a general solution. A particular solution specifically satisfies the non-homogeneous part of the differential equation. In essence, it matches the left-hand side of the equation to the right-hand side, providing a solution for just one instance.

Take, for example, the equation: \(y'' + y = e^x + x^2\). The right-hand side is \(e^x + x^2\), which hints at the kind of function we should consider for our particular solution. To build a particular solution, we start with a guessed form that mirrors the structure of the right-hand side of the equation, adding constants to be determined later.

The assumed form in this case is:

• \(y_p(x) = Ae^x + Bx^2 + Cx + D\)

We need variables like \(A\), \(B\), \(C\), and \(D\) because they are flexible and allow us to find the exact orientation of the solution. When we substitute this into the differential equation, we’ll later adjust these coefficients.

This method allows us to find a specific solution tailored to the input of the differential equation rather than the entire set of possible solutions.

Differential Equations

Differential equations play a crucial role in modeling real-world phenomena, describing how one or more functions change over time. These equations involve derivatives of a function or functions. Understanding and solving differential equations is critical for disciplines like physics, engineering, and even finance, as they describe systems ranging from planetary motion to electrical circuits.

A differential equation might be simple, like a first-order equation, or more complex. Often, they involve second-order derivatives, as seen with equations like \(y'' + y = e^x + x^2\) and others. That's why understanding how to manipulate these equations is vital to finding correct solutions.

In practice, solving such equations often involves:

• Identifying whether the equation is homogeneous or non-homogeneous.
• Determining the associated homogeneous solution.
• Using methods like undetermined coefficients to tackle the non-homogeneous part.

Using methods like undetermined coefficients requires formulating a hypothesis for the specific function that solves the equation. Thus, these equations are more than just numbers and variables; they are reflections of dynamic systems changing over time.

Coefficient Comparison

One of the vital skills in solving differential equations via the undetermined coefficients method is coefficient comparison. After assuming a form for the particular solution—like \(y_p(x) = Ae^x + Bx^2 + Cx + D\)—we substitute it into the differential equation. This is where coefficient comparison becomes critical.

We compare the coefficients from both sides of the equation. This step is how we solve for our unknown constants, such as \(A\), \(B\), \(C\), and \(D\).

Here’s how it works, broken down:

• First, ensure your equation is balanced in terms of similar terms (like \(e^x\) with \(e^x\), and \(x^2\) with \(x^2\)).
• Equate coefficients for each type of term.
• This results in a system of equations to solve for your coefficients.

For example, for the equation \(y'' + y = e^x + x^2\), through careful analysis, we deduce:

• \(A = 1\) by matching the coefficient of \(e^x\)
• \(B = 0\), \(C = 0\), \(D = 0\) by aligning other terms

Through this process, the seemingly complicated task of matching coefficients becomes manageable, allowing us to find the right particular solution effectively.