Solving 2y'' - 12y' + 20y = 0 General Solution And Coefficients

Hey guys! Let's dive into the fascinating world of differential equations. Today, we're tackling the equation 2y'' - 12y' + 20y = 0. Our mission? To find its general solution and figure out those coefficients, a and b, for some proposed functions. Buckle up; it's going to be an exciting ride!

Understanding the Differential Equation

Differential equations, like our 2y'' - 12y' + 20y = 0, are mathematical expressions that relate a function with its derivatives. In simpler terms, they describe how things change. This specific equation is a second-order linear homogeneous differential equation with constant coefficients. Sounds complicated? Don't worry; we'll break it down. The "second-order" part means the highest derivative involved is the second derivative (y''). "Linear" implies that the equation is linear in y and its derivatives. "Homogeneous" means there's no term without a y or its derivatives. And "constant coefficients" simply means the numbers multiplying y'' , y', and y are constants.

This type of equation pops up in various fields, from physics (think oscillations and vibrations) to engineering (circuit analysis) and even economics (modeling growth). So, understanding how to solve them is super important. To solve differential equations like this one, we usually guess a solution form and then plug it in to see if it works. Because our equation has constant coefficients, we'll try a solution of the form y = e^(rx), where r is a constant we need to find. This approach is based on the idea that exponential functions have derivatives that are proportional to themselves, which can lead to cancellations and simplifications in the equation. The key to solving these equations lies in finding the characteristic equation, solving for its roots, and then constructing the general solution based on the nature of these roots. This step-by-step process ensures that we capture all possible solutions and express them in a concise and usable form.

The Characteristic Equation: Our First Step

The secret weapon in solving this type of differential equation is the characteristic equation. We get this by replacing y'' with r^2, y' with r, and y with 1 in our original equation. So, 2y'' - 12y' + 20y = 0 transforms into:

2r^2 - 12r + 20 = 0

Now, we have a simple quadratic equation! We can simplify this by dividing everything by 2:

r^2 - 6r + 10 = 0

This quadratic equation is known as the characteristic equation, and its roots will dictate the form of our general solution. Solving the characteristic equation is a crucial step in finding the general solution of the differential equation. The roots of this equation provide the exponents for the exponential terms in the general solution. The method we use to solve the quadratic equation will depend on its form. In this case, since it doesn't factor easily, we will apply the quadratic formula to determine the roots. Once we have these roots, we can construct the general solution by combining exponential functions with these exponents, considering whether the roots are real, repeated, or complex. The characteristic equation is the bridge that connects the differential equation to its algebraic solution, making it a central concept in this process.

Solving the Quadratic Equation: Finding the Roots

Time to solve for r! This quadratic doesn't factor nicely, so we'll use the quadratic formula:

r = [-b ± √(b^2 - 4ac)] / (2a)

Where a = 1, b = -6, and c = 10. Plugging in these values, we get:

r = [6 ± √((-6)^2 - 4 * 1 * 10)] / (2 * 1) r = [6 ± √(36 - 40)] / 2 r = [6 ± √(-4)] / 2 r = [6 ± 2i] / 2 r = 3 ± i

Aha! We've got complex roots: r1 = 3 + i and r2 = 3 - i. Complex roots mean our solution will involve trigonometric functions. When the roots of the characteristic equation are complex, they come in conjugate pairs (like 3 + i and 3 - i). These complex roots indicate that the solutions to the differential equation will oscillate. The real part of the complex root (in this case, 3) corresponds to the exponential decay or growth, while the imaginary part (in this case, 1) corresponds to the frequency of oscillation. Understanding the nature of these roots is essential for constructing the correct form of the general solution. The presence of complex roots enriches the behavior of the solutions, allowing them to capture oscillatory phenomena, which are common in many physical systems modeled by differential equations.

Constructing the General Solution: Putting It All Together

Since we have complex roots of the form α ± βi (where α = 3 and β = 1), the general solution looks like this:

y(x) = e^(αx) (A cos(βx) + B sin(βx))

Plugging in our values for α and β, we get:

y(x) = e^(3x) (A cos(x) + B sin(x))

Here, A and B are arbitrary constants. This is the general solution to our differential equation. The general solution incorporates two arbitrary constants, A and B, because we started with a second-order differential equation, which requires two initial conditions to determine a unique solution. These constants allow the solution to fit a wide range of initial states. The exponential term e^(3x) indicates that the amplitude of the oscillations will grow over time since the real part of the complex root is positive. The cosine and sine terms, cos(x) and sin(x), reflect the oscillatory nature of the solution with a frequency determined by the imaginary part of the complex root. This general solution provides a comprehensive description of all possible behaviors of the system modeled by the differential equation, given different initial conditions.

Evaluating the Proposed Functions: Which One Fits?

Now, let's check the proposed functions. Our general solution is:

y(x) = e^(3x) (A cos(x) + B sin(x))

Comparing this to the options, we can see that:

• A) ae^(3x) cos(x) + be^(3x) sen(x) matches our general solution form.
• B) ae^(-3) cos(x) + be^(-3) sen(x) has the wrong exponential term (e^(-3) instead of e^(3x)).
• C) ae^(3z) cos(x) + be^(3) sen(x) also has incorrect exponential terms and mixes variables (z and x).

So, option A is the correct form for the solution. Comparing the proposed functions with the derived general solution is a crucial step to validate our result. The general solution provides the framework, and the specific form of the functions proposed should align with this framework. In this case, option A clearly matches the structure of our general solution, with the exponential term e^(3x) and the trigonometric terms cos(x) and sin(x). Options B and C deviate from this form, indicating that they are not valid solutions to the differential equation. This comparison reinforces the importance of understanding the components of the general solution and how they correspond to the roots of the characteristic equation.

Determining the Coefficients: A and B

The coefficients a and b (which are the same as our A and B) are arbitrary constants. To find specific values for them, we'd need initial conditions – values of y(x) and y'(x) at a particular point. Without initial conditions, we can't determine unique values for a and b; they remain arbitrary. Determining the coefficients A and B requires additional information, typically in the form of initial conditions. Initial conditions specify the value of the function and its derivative at a particular point, allowing us to solve for the unique values of A and B that satisfy these conditions. For example, if we knew y(0) and y'(0), we could substitute these values into the general solution and its derivative to obtain a system of two equations with two unknowns (A and B). Solving this system would give us the specific values of the coefficients, thus defining a unique solution to the differential equation. In the absence of initial conditions, A and B remain arbitrary constants, representing a family of solutions that all satisfy the differential equation.

Conclusion: Mastering Differential Equations

And there you have it! We've successfully found the general solution to 2y'' - 12y' + 20y = 0 and identified the correct form of the proposed functions. Differential equations might seem daunting at first, but with a step-by-step approach, they become much more manageable. Remember to find the characteristic equation, solve for the roots, and use those roots to build your general solution. Keep practicing, and you'll be a differential equation whiz in no time! Mastering differential equations is a valuable skill in many technical fields. The process we followed, from setting up the characteristic equation to constructing the general solution, illustrates a systematic approach that can be applied to a wide range of linear homogeneous differential equations with constant coefficients. Understanding the role of complex roots and how they lead to oscillatory solutions is particularly important. By practicing and applying these techniques, you can develop a strong foundation for tackling more complex problems in mathematical modeling and analysis. So, keep exploring, keep learning, and keep solving!