Problem 2 - Differential Equations and Linear Algebra

Question

For all problems below, use a complex-valued trial solution to determine a particular solution to the given differential equation. $$y^{\prime \prime}+2 y^{\prime}+y=50 \sin 3 x$$

Step-by-Step Solution

Answer

The particular solution to the given differential equation \(y'' + 2y' + y = 50\sin{3x}\) using the complex-valued trial solution method is \(y_p(x) = -\frac{25}{2}\sin{3x}\).

Step 1: Rewrite the non-homogeneous term using Euler's formula

Recall Euler's formula: \(e^{ix} = \cos x + i \sin x\). We will rewrite the non-homogeneous term using this formula: \(50\sin{3x} = 50i \cdot Im[e^{i3x}]\) So our differential equation becomes: \(y'' + 2y' + y = 50i \cdot Im[e^{i3x}]\)

Step 2: Find a trial solution

We consider a trial solution of the form: \(y_p(x) = (A + Bi)e^{i3x}\) Where A and B are unknown constants, and \(y_p(x)\) denotes a particular solution.

Step 3: Substitute the trial solution into the equation

Now, we find the first and second derivatives of the trial solution and substitute them into the differential equation: \(y'_p(x) = (A + Bi) \cdot 3i \cdot e^{i3x}\) \(y''_p(x) = (A + Bi) \cdot -9 \cdot e^{i3x}\) Substituting these into the differential equation, we get: \((A + Bi)(-9)e^{i3x} + 2(A + Bi)(3i)e^{i3x} + (A + Bi)e^{i3x} = 50i \cdot Im[e^{i3x}]\)

Step 4: Equate both sides of the equation and solve for the coefficients

Multiply like terms on the left-hand side of the equation: \(-(9A - 2A + A)e^{i3x} + i(-(9B - 6B + B)e^{i3x}) = 50i \cdot Im[e^{i3x}]\) Now we equate the coefficients of the imaginary parts on both sides and solve for A and B: \(-8A + 0i = 0\) \(-4B + 50i = 50i\) From the first equation, we get \(A = 0\). From the second equation, we get \(B = -\frac{25}{2}\).

Step 5: Write the particular solution

We substitute the values of A and B into the trial solution: \(y_p(x) = (A + Bi)e^{i3x} = (0 - \frac{25}{2}i)e^{i3x}\) Using Euler's formula, we find the imaginary part to obtain the final answer as \(y_p(x) = -\frac{25}{2}\sin{3x}\).

Key Concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. These equations play a crucial role in various fields such as physics, engineering, and economics.
They model the behavior of dynamic systems and can describe how quantities change over time.
There are two main types of differential equations: **ordinary differential equations (ODEs)** and **partial differential equations (PDEs)**.

• ODEs involve functions of a single variable and their derivatives.
• PDEs involve partial derivatives of functions with multiple variables.

In the exercise provided, the equation is an ordinary differential equation representing a second-order linear ODE:\[y'' + 2y' + y = 50 \sin{3x}\]This equation has a non-homogeneous component (the term on the right-hand side) \(50 \sin{3x}\), which requires finding a **particular solution** to solve it fully.

Euler's Formula

Euler's formula is a fundamental concept in complex analysis, linking trigonometry with exponential functions.It states that:\[e^{ix} = \cos x + i \sin x\]This elegant equation shows how exponential functions can express sinusoidal behavior using complex numbers.
In the context of solving differential equations, Euler’s formula simplifies the process of handling sine and cosine terms by using exponential representation.
In the provided exercise, we transformed the term \(50 \sin{3x}\) into a form involving Euler’s formula:\[50 \sin{3x} = 50i \cdot \text{Im}[e^{i3x}]\]This transformation allowed us to consider a **complex-valued trial solution**, simplifying working with derivatives and integration in a unified framework.

Particular Solution

Finding a particular solution means identifying a specific solution to the differential equation that satisfies the non-homogeneous part.
In this exercise, we assumed our solution in the form of a **trial solution** due to the non-homogeneous term \(50 \sin{3x}\).
The trial solution form was:\[y_p(x) = (A + Bi)e^{i3x}\]Here, \(A\) and \(B\) are constants to be determined by substituting back into the original differential equation and equating the like terms.
The real and imaginary parts are then aligned with those on the equation's right side to extract these constants.As calculated:

• \(A = 0\)
• \(B = -\frac{25}{2}\)

Substituting these values back gives us the particular solution:\[y_p(x) = (0 - \frac{25}{2}i)e^{i3x}\]Finally, using Euler's formula, the imaginary part provides the solution: \[y_p(x) = -\frac{25}{2} \sin{3x}\]This particular solution represents the specific function satisfying the original differential equation.