Problem 15 - 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 \( y(t) = C_1 \cos(\sqrt{5}t) + C_2 \sin(\sqrt{5}t) + 8 \sin(2t) \).

Step 1: Identify the Differential Equation Type

The given equation is a second-order linear differential equation of the form:\[ y'' + 5y = 8 \sin(2t) \]Here, \(y''\) is the second derivative of \(y\) with respect to \(t\). This is a non-homogeneous differential equation.

Step 2: Solve the Homogeneous Equation

Consider the associated homogeneous equation:\[ y'' + 5y = 0 \]To solve it, assume a solution of the form \(y = e^{rt}\). Substituting this into the equation gives:\[ r^2 e^{rt} + 5e^{rt} = 0 \]Factor out \(e^{rt}\) and set the characteristic equation: \[ r^2 + 5 = 0 \]Solving for \(r\), we get:\[ r = \pm i\sqrt{5} \]The general solution to the homogeneous equation is: \[ y_h(t) = C_1 \cos(\sqrt{5}t) + C_2 \sin(\sqrt{5}t) \]

Step 3: Find a Particular Solution

For the non-homogeneous part \(8 \sin(2t)\), use the method of undetermined coefficients. Assume a particular solution of the form:\[ y_p(t) = A \sin(2t) + B \cos(2t) \]Differentiate \(y_p(t)\):\[ y_p'(t) = 2A \cos(2t) - 2B \sin(2t) \]\[ y_p''(t) = -4A \sin(2t) - 4B \cos(2t) \]Substitute these into the original equation:\[-4A \sin(2t) - 4B \cos(2t) + 5(A \sin(2t) + B \cos(2t)) = 8\sin(2t) \]Simplify and collect terms:\[ (5A - 4A) \sin(2t) + (5B - 4B) \cos(2t) = 8\sin(2t) \]\[ A \sin(2t) + B \cos(2t) = 8 \sin(2t) \]Equate coefficients: \(A = 8\) and \(B = 0\).Thus, the particular solution is:\[ y_p(t) = 8 \sin(2t) \]

Step 4: Construct the General Solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions:\[ y(t) = y_h(t) + y_p(t) \]\[ y(t) = C_1 \cos(\sqrt{5}t) + C_2 \sin(\sqrt{5}t) + 8 \sin(2t) \]

Step 5: Conclusion

The general solution to the differential equation \( y'' + 5y = 8 \sin(2t) \) is:\[ y(t) = C_1 \cos(\sqrt{5}t) + C_2 \sin(\sqrt{5}t) + 8 \sin(2t) \]

Key Concepts

Homogeneous Equation

A homogeneous equation is a differential equation in which every term is a function of the dependent variable and its derivatives. For example, in the solution to the equation \( y'' + 5y = 8 \sin(2t) \), the associated homogeneous equation is \( y'' + 5y = 0 \).
This means the equation only contains terms involving \( y \) and its derivatives, without any external forcing functions like \( 8 \sin(2t) \).

To solve a homogeneous equation, you typically start by assuming a solution of the form \( y = e^{rt} \). This leads to the characteristic equation, which helps determine the general solution for \( y \).

• The characteristic equation for \( y'' + 5y = 0 \) is \( r^2 + 5 = 0 \).
• Solving for \( r \), you obtain complex roots \( r = \pm i\sqrt{5} \).

This indicates oscillatory solutions, expressed as linear combinations of sine and cosine functions. The solution for the homogeneous part becomes \( y_h(t) = C_1 \cos(\sqrt{5}t) + C_2 \sin(\sqrt{5}t) \).
These terms form the complementary (or homogeneous) solution, capturing the behavior of the system without external forces.

Particular Solution

In differential equations, finding a particular solution involves determining a function that satisfies the non-homogeneous part of the equation. For the equation \( y'' + 5y = 8 \sin(2t) \), the external force is \( 8 \sin(2t) \).
For this, we assume a function form that resembles \( 8 \sin(2t) \), like \( y_p(t) = A \sin(2t) + B \cos(2t) \).

The next steps involve:

• Finding the derivatives of your assumed particular solution.
• Substituting these into the original differential equation.

This substitution leads to an equation where terms can be equated separately. After simplification, coefficients of like terms (\( \sin(2t) \) and \( \cos(2t) \)) help to find values for \( A \) and \( B \).
For \( y'' + 5y = 8 \sin(2t) \), this process gives:

• \( A = 8 \)
• \( B = 0 \)

Thus, the particular solution is \( y_p(t) = 8 \sin(2t) \). This represents the influence of the external force on the system.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find the particular solution of linear differential equations with constant coefficients.
This method works well when the non-homogeneous part of the differential equation is a standard function like polynomials, exponentials, sines, or cosines.

Here's how it works:

• Assume a form for the particular solution based on the type of function on the right side of the equation.
• The form should include coefficients that are initially unknown, which are the 'undetermined coefficients'.
• Differentiate this assumed solution as needed and substitute back into the original differential equation.
• Gather like terms, solve the resulting system of equations, and identify the values of your assumed coefficients.

For instance, in our problem:

• The non-homogeneous term \( 8 \sin(2t) \) suggests trying a combination of sine and cosine in the assumed solution.

Thus, you introduce \( y_p(t) = A \sin(2t) + B \cos(2t) \), compute necessary derivatives, substitute, and solve for \( A \) and \( B \).
This method provides a systematic way to address non-homogeneous differential equations, streamlining the process of finding particular solutions.