Solve The Initial-value Problem: Y" - 4y' + 8y = 0, Y(0) = 1, Y'(0) = 2.

Solve The Initial-value Problem: Y" - 4y' + 8y = 0, Y(0) = 1, Y'(0) = 2.

Understanding how to solve differential equations with initial conditions is a fundamental aspect of applied mathematics and engineering. In this article, we focus on solving the initial-value problem (IVP):

Y'' - 4Y' + 8Y = 0, with initial conditions Y(0) = 1 and Y'(0) = 2.

This type of differential equation is a second-order linear homogeneous differential equation with constant coefficients. Solving it involves several steps, including finding the characteristic equation, determining the general solution, and applying the initial conditions to find specific constants.

This comprehensive guide will walk you through each step in detail, providing insights into the theory and practical methods used. Whether you are a student or a professional, understanding this process enhances your ability to tackle similar differential equations.

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Understanding the Differential Equation

The differential equation under consideration is:

Y'' - 4Y' + 8Y = 0

• Order: Second-order because the highest derivative is Y''.


• Linearity: Linear due to the sum of derivatives and the function itself, with constant coefficients.


• Homogeneity: No term involving Y explicitly multiplied by a function of x; the right-hand side is zero.

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Step 1: Write the Characteristic Equation

The method for solving linear homogeneous equations with constant coefficients involves assuming a solution of the form:

Y(x) = e^{rx}

where r is a constant to be determined.

Plugging Y = e^{rx} into the differential equation gives:

r^2 e^{rx} - 4r e^{rx} + 8 e^{rx} = 0

Dividing through by e^{rx} (which is never zero), we obtain the characteristic equation:

r^2 - 4r + 8 = 0

This quadratic equation determines the roots that will define the general solution.

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Step 2: Find the Roots of the Characteristic Equation

Solve the quadratic:

r^2 - 4r + 8 = 0

Using the quadratic formula:

r = [4 ± √(16 - 418)] / 2

Calculate the discriminant:

Δ = 16 - 32 = -16

Since the discriminant is negative, the roots are complex conjugates:

r = [4 ± √(-16)] / 2

r = [4 ± 4i] / 2

Simplify:

r = 2 ± 2i

These roots are complex conjugates with real part 2 and imaginary part 2.

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Step 3: Write the General Solution

For complex roots r = α ± βi, the general solution of the differential equation is:

Y(x) = e^{αx} [C1 cos(βx) + C2 sin(βx)]

In our case:

α = 2

β = 2

Therefore, the general solution is:

Y(x) = e^{2x} [C1 cos(2x) + C2 sin(2x)]

where C1 and C2 are arbitrary constants determined by initial conditions.

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Step 4: Apply Initial Conditions to Find Constants

Given initial conditions:

Y(0) = 1

Y'(0) = 2

Let's compute Y(0):

Y(0) = e^{0} [C1 cos(0) + C2 sin(0)] = 1 [C1 1 + C2 0] = C_1

So:

C_1 = 1

Next, compute Y'(x):

Y'(x) = derivative of Y(x)

Using the product rule:

Y'(x) = derivative of e^{2x} [C1 cos(2x) + C2 sin(2x)]

Y'(x) = e^{2x} 2 [C1 cos(2x) + C2 sin(2x)] + e^{2x} [ -2 C1 sin(2x) + 2 C2 cos(2x)]

Simplify:

Y'(x) = e^{2x} [ 2 C1 cos(2x) + 2 C2 sin(2x) - 2 C1 sin(2x) + 2 C2 cos(2x) ]

Combine like terms:

Y'(x) = e^{2x} [ (2 C1 cos(2x) + 2 C2 cos(2x)) + (2 C2 sin(2x) - 2 C1 sin(2x)) ]

Y'(x) = e^{2x} [ 2 (C1 + C2) cos(2x) + 2 (C2 - C1) sin(2x) ]

Evaluate at x=0:

Y'(0) = e^{0} [ 2 (C1 + C2) cos(0) + 2 (C2 - C1) sin(0) ] = 1 [ 2 (C1 + C2) 1 + 2 (C2 - C1) 0 ] = 2 (C1 + C2)

Given Y'(0) = 2, and C_1 = 1:

2 (1 + C_2) = 2

Divide both sides by 2:

1 + C_2 = 1

Solve for C_2:

C_2 = 0

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Final Solution

Substitute the constants back into the general solution:

Y(x) = e^{2x} [C1 cos(2x) + C2 sin(2x)] = e^{2x} [1 cos(2x) + 0 sin(2x)] = e^{2x} cos(2x)

Thus, the unique solution to the initial-value problem is:

Y(x) = e^{2x} cos(2x)

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Additional Insights and Applications

Understanding this solution provides valuable insights into the behavior of systems modeled by such differential equations:

• Oscillatory Growth: The exponential term e^{2x} indicates an exponential growth, while cosine introduces oscillations.


• Physical Systems: This solution can model phenomena like damped or amplified oscillations in mechanical or electrical systems.


• Stability: The positive real part in the exponential indicates the solution grows without bound as x increases, implying instability in some contexts.

Summary of Key Steps

To summarize, solving the initial-value problem involved:

1. Formulating the characteristic equation: r^2 - 4r + 8 = 0


2. Finding roots: Complex conjugates r = 2 ± 2i


3. Writing the general solution: Y(x) = e^{2x} [C1 cos(2x) + C2 sin(2x)]


4. Applying initial conditions: Y(0) = 1 and Y'(0) = 2, to find C1 = 1, C2 = 0


5. Final solution: Y(x) = e^{2x} cos(2x)

Conclusion

Solving second-order linear homogeneous differential equations with constant coefficients and initial conditions is a systematic process rooted in characteristic equations and exponential-trigonometric solutions. This example demonstrates how to approach such problems methodically, interpret the roots of the characteristic equation, and use initial conditions to determine specific solutions.

Mastering these techniques is essential for students and professionals working in engineering, physics, and applied mathematics, where such equations frequently model real-world systems. Practice with different types of equations and initial conditions will deepen your understanding and proficiency in differential equations.

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FAQs

Q1: What happens if the roots of the characteristic equation are real and repeated?
Answer: The general solution involves terms like e^{rx} and x e^{rx} to account for multiplicity.

Q2: How can I verify that my solution satisfies the initial conditions?
Answer: Substitute x=0 into your solution and its derivative; verify that they match the initial values.

Q3: Can this method be applied to non-homogeneous equations?
Answer: Yes, but it involves finding a particular solution in addition to the homogeneous solution.

Q4: Are there software tools to help solve such differential equations?
Answer: Yes, tools like WolframAlpha, MATLAB, Mathematica, and online calculators can assist in solving and verifying solutions.

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Keywords: differential equations, initial-value problem, second-order linear differential equation, characteristic equation, complex roots, exponential oscillations, mathematical modeling, solution methods