Problem 27 - Differential Equations and Linear Algebra

Question

Determine the general solution to the given differential equation. Derive your trial solution using the annihilator technique. $$y^{\prime \prime}+4 y=8 \sin 2 x$$.

Step-by-Step Solution

Answer

The general solution to the given differential equation \(y^{\prime\prime}+4y=8\sin(2x)\) using the annihilator technique is \(y(x) = C_1\cos(2x) + C_2\sin(2x)\), where \(C_1\) and \(C_2\) are constants.

Step 1: Determine the auxiliary equation and its roots

The auxiliary equation for the homogeneous equation \(y''+4y=0\) is \[m^2 + 4 = 0\] Solving for \(m\) gives the roots \(m = \pm 2i\).

Step 2: Find the complementary solution

The complementary solution is given by the general solution of the homogeneous equation, which is of the form: \[y_c(x) = C_1 \cos(2x) + C_2 \sin(2x)\] where \(C_1\) and \(C_2\) are constants.

Step 3: Apply the annihilator operator on the inhomogeneous DE

To do this, first we need to determine the appropriate annihilator operator that will annihilate the right-hand side of the given DE. Since the right-hand side is of the form \(8\sin(2x)\), the corresponding annihilator operator is \((D^2 + 4)\), where \(D\equiv\frac{d}{dx}\). Apply the annihilator operator on both sides of the given DE: \[(D^2+4)(y''+4y) = (D^2+4)(8\sin(2x))\] The left-hand side simplifies to \(y^{(4)}+8y''\), and the right-hand side simplifies to \(0\), because the annihilator operator annihilates the term it is applied on. Therefore, our new DE is: \[y^{(4)}+8y'' = 0\]

Step 4: Find the particular solution

To find the particular solution for this differential equation, let's first solve its auxiliary equation: \[m^4 + 8m^2 = 0\] The roots of this polynomial are \(m=0, 0, + 2i, - 2i\). So, the general solution for the fourth-order homogeneous equation is: \[y_h(x) = A_1 + A_2 x + A_3 \cos(2x) + A_4 \sin(2x)\] Our trial solution must not contain any terms from the complementary solution we found in Step 2. Thus, the trial solution would be of the form: \[y_p(x) = B_1 + B_2 x\] Substitute this trial solution into the given DE: \[((B_1 + B_2 x)'' + 4(B_1 + B_2 x) = 8\sin(2x)\] Simplify the equation: \[4(B_1 + B_2 x) = 8 \sin(2x)\] We can see that there are no constants or linear terms on the right-hand side of the equation, so we conclude that \(B_1\) and \(B_2\) must be equal to zero. Therefore, there is no particular solution for this DE.

Step 5: Combine complementary and particular solutions

Since there's no particular solution, the general solution of the DE is just the complementary solution: \[y(x) = y_c(x) = C_1 \cos(2x) + C_2 \sin(2x)\] This is the general solution to the given differential equation.

Key Concepts

Differential Equation

A differential equation is a mathematical equation that involves functions and their derivatives. It serves as a tool for describing various phenomena in physics, engineering, biology, and more.
In our example, the differential equation given is \( y'' + 4y = 8 \sin 2x \). It involves the second derivative \( y'' \) of a function \( y \), and it includes a simple polynomial term \( 4y \).
The equation is non-homogeneous because it includes a term (the \( 8 \sin 2x \)) that does not contain the dependent variable \( y \) or its derivatives. To solve it, we need to find a function \( y(x) \) that satisfies the equation for all \( x \) values in its domain.

Complementary Solution

The complementary solution represents the general solution to the homogeneous part of a differential equation. It contains all solutions to the equation when the non-homogeneous term is zero.
In our problem, the complementary solution stems from solving \( y'' + 4y = 0 \), which is the homogeneous version of the given equation. Solving this involves finding the roots of the auxiliary equation \( m^2 + 4 = 0 \).
These roots are complex, \( m = \pm 2i \), leading to a complementary solution:

• \( y_c(x) = C_1 \cos(2x) + C_2 \sin(2x) \)

where \( C_1 \) and \( C_2 \) are arbitrary constants. This solution covers all functions that solve the differential equation without considering the non-homogeneous part (i.e., the \( 8 \sin 2x \)).

Auxiliary Equation

The auxiliary equation is a polynomial equation derived from the homogeneous part of a differential equation, which helps find the complementary solution.
Here, for the homogeneous equation \( y'' + 4y = 0 \), the process involves replacing \( y'' \) and \( y \) with powers of \( m \), resulting in the auxiliary equation \( m^2 + 4 = 0 \).
Solving for \( m \) involves finding the roots, which usually includes:

• Real roots that define exponential functions in the solution
• Complex roots that give rise to trigonometric (sine and cosine) functions

For this equation, the complex roots \( \pm 2i \) lead us to the complementary solution involving sine and cosine functions, as described earlier. The auxiliary equation essentially bridges the homogeneous equation and its solutions.

Particular Solution

The particular solution aims to find a specific function that satisfies the non-homogeneous differential equation. It accounts for the influence of the non-homogeneous term, leading us to the complete general solution.
In our problem, we must consider the term \( 8 \sin 2x \). The annihilator technique uses an operator that turns this term into zero (essentially annulling it), simplifying the solving process.
We applied the operator \( (D^2 + 4) \) to the entire equation. This led to a new equation \( y^{(4)} + 8y'' = 0 \) but ultimately revealed that our trial solution \( y_p(x) = B_1 + B_2 x \) does not contribute any non-zero terms suitable for cancellation, indicating no particular solution.
Therefore, the general solution heavily relies on the complementary solution alone, embodying the homogeneous root structure.