Problem 14 - Calculus: Early Transcendentals

Question

Find the general solution to the differential equation. $$y^{\prime \prime}+5 y=8 \sin (2 t)$$

Step-by-Step Solution

Answer

The general solution is \(y = C_1\cos(\sqrt{5}t) + C_2\sin(\sqrt{5}t) + 8\sin(2t)\).

Step 1: Rewrite the Differential Equation

We start by recognizing the given differential equation: \[y'' + 5y = 8 \sin(2t)\].This is a second-order linear non-homogeneous differential equation.

Step 2: Solve the Homogeneous Equation

Find the complementary solution \(y_c\) by solving the homogeneous equation:\[y'' + 5y = 0\].Assuming a solution of the form \(y = e^{rt}\), we substitute into the equation to get the characteristic equation:\[r^2 + 5 = 0\].Solving, \(r^2 = -5\) which gives the roots \(r = \pm \sqrt{5}i\).Thus, the complementary solution is \[y_c = C_1\cos(\sqrt{5}t) + C_2\sin(\sqrt{5}t)\].

Step 3: Solve for the Particular Solution

To find the particular solution \(y_p\), we use the method of undetermined coefficients based on the non-homogeneous term \(8\sin(2t)\).Assume a solution of the form:\[y_p = A\sin(2t) + B\cos(2t)\].Compute the first and second derivatives:\[y_p' = 2A \cos(2t) - 2B \sin(2t)\]\[y_p'' = -4A \sin(2t) - 4B \cos(2t)\].Substitute \(y_p\) and its derivatives into the original differential equation:\[-4A \sin(2t) - 4B \cos(2t) + 5(A \sin(2t) + B \cos(2t)) = 8 \sin(2t)\].Simplifying gives:\[A\sin(2t) + B\cos(2t) = 8\sin(2t)\].Matching coefficients, \(A = 8\) and \(B = 0\).So, \(y_p = 8\sin(2t)\).

Step 4: Write the General Solution

The general solution is the sum of the complementary and particular solutions:\[y = y_c + y_p\].Substitute both solutions found:\[y = C_1\cos(\sqrt{5}t) + C_2\sin(\sqrt{5}t) + 8\sin(2t)\].

Key Concepts

Complementary Solution

When dealing with second-order non-homogeneous differential equations like \[ y'' + 5y = 8 \sin(2t) \],it's often helpful to first look for a solution to the homogeneous version of the equation. This part of the solution is known as the complementary solution.

The homogeneous version is derived by simply removing the non-homogeneous term, transforming our given equation to\[ y'' + 5y = 0 \].Assuming solutions of the form \( y = e^{rt} \), we substitute \( y \) and its derivatives back into the equation. This results in a characteristic equation,

• \( r^2 + 5 = 0 \)

giving us complex conjugate roots \( r = \pm \sqrt{5}i \).

This leads to the complementary solution:\[ y_c = C_1\cos(\sqrt{5}t) + C_2\sin(\sqrt{5}t) \].Here, \( C_1 \) and \( C_2 \) are arbitrary constants determined by the initial conditions. These harmonic functions arise due to the imaginary parts of the roots, indicating an oscillatory behavior typical of such systems.

Particular Solution

The particular solution, denoted by \( y_p \), is the part of the solution that "addresses" the non-homogeneous aspect of the differential equation.

In our exercise with the equation \[ y'' + 5y = 8 \sin(2t) \],the particular solution is derived to fit the right-hand side, which is \( 8\sin(2t) \).

To determine \( y_p \), we propose a form similar to the non-homogeneous term. For \( 8\sin(2t) \), an appropriate form to guess would be

• \( y_p = A\sin(2t) + B\cos(2t) \)

Next, it's time for some calculus! By calculating the first and second derivatives of \( y_p \) and plugging them back into the left side of the original differential equation, we can match coefficients with those on the right side.

Through this process, we find:

• \( A = 8 \)
• \( B = 0 \)

Thus, the particular solution becomes\[ y_p = 8 \sin(2t) \].This approach ensures that \( y_p \) complements the non-homogeneous term when the equation is solved.

Method of Undetermined Coefficients

One effective technique for solving non-homogeneous linear differential equations is the method of undetermined coefficients. This approach is especially useful when the non-homogeneous term is a common function like polynomials, exponentials, sines, or cosines.

The core idea involves assuming a form for the particular solution based on the non-homogeneous component. This assumed solution will contain constants (coefficients) that we initially don't know.

In the differential equation \[ y'' + 5y = 8 \sin(2t) \],we used the method of undetermined coefficients by assuming a solution format

• \( y_p = A\sin(2t) + B\cos(2t) \)

because our non-homogeneous part is \( 8\sin(2t) \).

The next step requires substituting \( y_p \) and its derivatives back into the differential equation, then adjusting \( A \) and \( B \) to fulfill the equation's requirements. This process involves comparing terms of the same type (sine or cosine) on both sides of the equation to determine the coefficients.

Through the example we've explored, as we calculated earlier:

• \( A = 8 \)
• \( B = 0 \)

These coefficients complete our particular solution and allow us to find an overall general solution for the equation.