4. Find The Solution To The Differential Equation Y"(t) + 5y'(t) + 2y(t) = 3u(t), Where Y(0) = A And

Understanding the Differential Equation: Y"(t) + 5y'(t) + 2y(t) = 3u(t), Where Y(0) = A

4. Find The Solution To The Differential Equation Y"(t) + 5y'(t) + 2y(t) = 3u(t), Where Y(0) = A And is a classic problem in differential equations, often encountered in engineering and physics. This type of second-order linear differential equation with constant coefficients models a variety of physical systems, such as mechanical vibrations, electrical circuits, and control systems.

In this article, we'll explore the steps to solve this differential equation comprehensively. Whether you're a student, educator, or professional, understanding the solution process will deepen your grasp of differential equations and their applications.

Overview of the Differential Equation Components

Understanding the Terms

• Y"(t): The second derivative of y(t), representing acceleration or curvature depending on the context.
• Y'(t): The first derivative of y(t), reflecting velocity or rate of change.
• y(t): The function we aim to find.
• 3u(t): The forcing function, where u(t) is the unit step function (Heaviside function). This indicates a sudden change or input at t=0.
• Y(0) = A: The initial condition specifying the value of y(t) at t=0.

Implications of the Forcing Function

The presence of the unit step function u(t) makes the differential equation a nonhomogeneous equation. Its effect is to introduce a constant input starting at t=0, which influences the particular solution.

Step 1: Formulate the Homogeneous Equation

To find the general solution, begin by solving the homogeneous differential equation:

Y"(t) + 5Y'(t) + 2Y(t) = 0

This homogeneous equation reflects the system's natural response without external forcing.

Step 2: Solve the Characteristic Equation

Deriving the Characteristic Equation

Assuming solutions of the form y(t) = e^{rt}, substitute into the homogeneous differential equation:

r^2 e^{rt} + 5r e^{rt} + 2 e^{rt} = 0

Dividing through by e^{rt} (which is never zero), we get:

r^2 + 5r + 2 = 0

Solving for r

Apply the quadratic formula:

r = \frac{-5 \pm \sqrt{(5)^2 - 4 \times 1 \times 2}}{2}

r = \frac{-5 \pm \sqrt{25 - 8}}{2} = \frac{-5 \pm \sqrt{17}}{2}

Roots of the Characteristic Equation

• r₁ = \(\frac{-5 + \sqrt{17}}{2}\)
• r₂ = \(\frac{-5 - \sqrt{17}}{2}\)

Step 3: Write the General Homogeneous Solution

The general solution to the homogeneous equation is:

yh(t) = C1 e^{r1 t} + C2 e^{r2 t}

where C₁ and C₂ are constants determined by initial conditions.

Step 4: Find the Particular Solution

Addressing the Nonhomogeneous Term

The nonhomogeneous term is 3u(t), which is a constant input for t ≥ 0. To find the particular solution, y_p(t), we typically assume a form based on the forcing function.

Assumption for the Particular Solution

• If the forcing term is a constant, a good initial guess is y_p(t) = K, a constant.
• Alternatively, for differential equations, a constant particular solution is often effective when the right side is a constant.

Determine K by Substitution

Assuming y_p(t) = K, then:

Y'(t) = 0, \quad Y''(t) = 0

Substitute into the differential equation:

0 + 50 + 2K = 3u(t)

For t ≥ 0, u(t) = 1, so:

2K = 3

K = \frac{3}{2}

Step 5: Write the General Solution

The overall general solution combines the homogeneous and particular solutions:

y(t) = yh(t) + yp(t) = C1 e^{r1 t} + C2 e^{r2 t} + \frac{3}{2}

Step 6: Apply Initial Conditions to Determine Constants

Using Y(0) = A

At t=0, the solution becomes:

A = C1 e^{r1 \times 0} + C2 e^{r2 \times 0} + \frac{3}{2}

A = C1 + C2 + \frac{3}{2}

Find Y'(t) for initial velocity condition (if given)

In many problems, an initial velocity condition Y'(0) = B is provided. If so, differentiate the general solution:

Y'(t) = C1 r1 e^{r1 t} + C2 r2 e^{r2 t}

At t=0:

Y'(0) = C1 r1 + C2 r2

Using the initial velocity condition Y'(0) = B, solve for C₁ and C₂.

Summary of the Solution Process

• Formulate the homogeneous differential equation and solve its characteristic equation.
• Write the homogeneous solution based on the roots.
• Identify the nonhomogeneous term and assume a particular solution accordingly.
• Calculate the particular solution, often a constant in this case.
• Combine the homogeneous and particular solutions to obtain the general solution.
• Apply initial conditions to solve for constants.

Additional Considerations

Handling Discontinuities and Step Inputs

The presence of the unit step function u(t) indicates that the input begins at t=0. This often leads to a discontinuity or a sudden change in the system's response. To handle such cases, the Laplace Transform method is highly effective.

Laplace Transform Method

Applying the Laplace transform converts the differential equation into an algebraic equation, simplifying the process of solving for y(t). Here’s a brief outline:

• Take the Laplace transform of each term, considering initial conditions.
• Solve the algebraic equation for Y(s), the Laplace transform of y(t).
• Use inverse Laplace transform to find y(t).

Using Laplace Transform to Solve the Differential Equation

Transform the Equation

L{Y"(t)} + 5 L{Y'(t)} + 2 L{Y(t)} = 3 L{u(t)}

Recall that:

• L{Y"(t)}