Solving Differential Equations: A Step-by-Step Guide

Hey guys! Today, we're diving into the exciting world of differential equations. Don't worry, it's not as scary as it sounds! We'll break it down step by step, so you can tackle these problems like a pro. We've got two equations to solve, each with its own set of initial conditions. Let's get started!

1. Solving $y'' - 4y' + 3y = 0$ with $y(0) = 4$ and $y'(0) = 2$

Okay, so the first equation we're going to tackle is $y'' - 4y' + 3y = 0$. This is a second-order linear homogeneous differential equation with constant coefficients. That's a mouthful, I know, but it just means we can solve it using a standard method. So, let's see how to do it:

Finding the General Solution

First, we assume a solution of the form $y = e^{rx}$, where $r$ is a constant. This is a common trick for solving these types of equations. Now, we need to find the first and second derivatives of $y$:

• $y' = re^{rx}$
• $y'' = r^2e^{rx}$

Next, substitute these into the original differential equation:

$r^2e^{rx} - 4re^{rx} + 3e^{rx} = 0$

We can factor out $e^{rx}$ since it's a common term:

$e^{rx}(r^2 - 4r + 3) = 0$

Since $e^{rx}$ is never zero, we can focus on the quadratic equation:

$r^2 - 4r + 3 = 0$

This is called the characteristic equation. We can factor this quadratic equation to find the roots:

$(r - 1)(r - 3) = 0$

So, the roots are $r_1 = 1$ and $r_2 = 3$. Since we have two distinct real roots, the general solution is given by:

$y(x) = c_1e^{x} + c_2e^{3x}$

where $c_1$ and $c_2$ are arbitrary constants. This is the general solution to the differential equation.

Applying Initial Conditions

Now, we need to find the particular solution that satisfies the given initial conditions: $y(0) = 4$ and $y'(0) = 2$. First, let's use the condition $y(0) = 4$:

$y(0) = c_1e^{0} + c_2e^{3(0)} = c_1 + c_2 = 4$

So, we have our first equation:

$c_1 + c_2 = 4$

Next, we need to find the derivative of the general solution:

$y'(x) = c_1e^{x} + 3c_2e^{3x}$

Now, apply the second initial condition $y'(0) = 2$:

$y'(0) = c_1e^{0} + 3c_2e^{3(0)} = c_1 + 3c_2 = 2$

So, we have our second equation:

$c_1 + 3c_2 = 2$

Now, we have a system of two linear equations with two unknowns:

1. $c_1 + c_2 = 4$
2. $c_1 + 3c_2 = 2$

We can solve this system by subtracting the first equation from the second equation:

$(c_1 + 3c_2) - (c_1 + c_2) = 2 - 4$

$2c_2 = -2$

$c_2 = -1$

Now, substitute $c_2 = -1$ into the first equation:

$c_1 + (-1) = 4$

$c_1 = 5$

So, we have found the values of the constants: $c_1 = 5$ and $c_2 = -1$.

The Particular Solution

Finally, substitute the values of $c_1$ and $c_2$ into the general solution to get the particular solution:

$y(x) = 5e^{x} - e^{3x}$

This is the particular solution that satisfies the given differential equation and initial conditions. Great job, we solved the first one!

2. Solving $y'' + 4y = 0$ with $y(0) = 10$ and $y'(0) = 8$

Now, let's move on to the second equation: $y'' + 4y = 0$. This is another second-order linear homogeneous differential equation with constant coefficients, but this time, it's a bit different because it has no $y'$ term. This will lead to complex roots in the characteristic equation.

Finding the General Solution

As before, we assume a solution of the form $y = e^{rx}$. Then we find the first and second derivatives:

• $y' = re^{rx}$
• $y'' = r^2e^{rx}$

Substitute these into the original differential equation:

$r^2e^{rx} + 4e^{rx} = 0$

Factor out $e^{rx}$:

$e^{rx}(r^2 + 4) = 0$

Since $e^{rx}$ is never zero, we focus on the characteristic equation:

$r^2 + 4 = 0$

Solve for $r$:

$r^2 = -4$

$r = \{\pm 2i\}$

Since we have complex roots $r = \{\alpha \pm i\beta\}$ where $\alpha = 0$ and $\beta = 2$, the general solution is given by:

$y(x) = c_1\cos(2x) + c_2\sin(2x)$

where $c_1$ and $c_2$ are arbitrary constants. This is the general solution for the second differential equation.

Applying Initial Conditions

Now, we need to apply the initial conditions $y(0) = 10$ and $y'(0) = 8$ to find the particular solution. First, let's use the condition $y(0) = 10$:

$y(0) = c_1\cos(2(0)) + c_2\sin(2(0)) = c_1\cos(0) + c_2\sin(0) = c_1(1) + c_2(0) = c_1 = 10$

So, we have found $c_1 = 10$.

Next, we need to find the derivative of the general solution:

$y'(x) = -2c_1\sin(2x) + 2c_2\cos(2x)$

Now, apply the second initial condition $y'(0) = 8$:

$y'(0) = -2c_1\sin(2(0)) + 2c_2\cos(2(0)) = -2c_1\sin(0) + 2c_2\cos(0) = -2c_1(0) + 2c_2(1) = 2c_2 = 8$

So, we have:

$2c_2 = 8$

$c_2 = 4$

Thus, we have found the values of the constants: $c_1 = 10$ and $c_2 = 4$.

The Particular Solution

Finally, substitute the values of $c_1$ and $c_2$ into the general solution to get the particular solution:

$y(x) = 10\cos(2x) + 4\sin(2x)$

This is the particular solution that satisfies the given differential equation and initial conditions. Awesome, we solved the second one too!

Conclusion

So, there you have it! We've successfully solved two different types of second-order linear homogeneous differential equations with constant coefficients. Remember, the key steps are:

1. Find the general solution by assuming a solution of the form $y = e^{rx}$ and solving the characteristic equation.
2. Apply the initial conditions to find the values of the arbitrary constants.
3. Substitute the values of the constants back into the general solution to get the particular solution.

Keep practicing, and you'll become a differential equation master in no time! Good luck, and have fun solving more problems! Differential equations are incredibly useful in modeling real-world phenomena, ranging from the motion of a pendulum to the spread of diseases. The process of finding both general and particular solutions equips you with a powerful toolkit for understanding and predicting the behavior of dynamic systems. Mastering these techniques will open doors to more advanced topics in mathematics, physics, engineering, and other scientific disciplines. Remember that each problem might require slightly different approaches, so it's essential to develop a strong understanding of the underlying principles.

Furthermore, consider exploring numerical methods for solving differential equations, especially when analytical solutions are difficult or impossible to find. Methods like Euler's method or Runge-Kutta methods can provide accurate approximations of the solutions, allowing you to gain insights into the behavior of complex systems. These numerical techniques are often implemented using computer software, making them invaluable tools for researchers and practitioners. So, delve deeper, experiment with different methods, and embrace the beauty and power of differential equations in describing the world around us. The journey might be challenging at times, but the rewards of understanding and applying these concepts are immense. Keep pushing yourself, and you'll be amazed at what you can achieve.

Differential equations are a fundamental part of mathematical analysis and have a vast array of applications in various fields. Understanding how to solve them is crucial for anyone pursuing a career in science, engineering, or mathematics. The two examples we've worked through illustrate the basic techniques involved, but there are many more advanced methods to explore. For instance, you might encounter non-homogeneous equations, where the right-hand side is not zero. These require different approaches, such as the method of undetermined coefficients or variation of parameters. Each method has its own strengths and weaknesses, and the choice of which one to use depends on the specific form of the equation. Additionally, you'll encounter systems of differential equations, where multiple equations are coupled together. These can be more challenging to solve, but they are essential for modeling complex systems with interacting components. The world of differential equations is vast and fascinating, so keep exploring and expanding your knowledge. The more you learn, the better equipped you'll be to tackle real-world problems and make meaningful contributions to your field. Remember to always practice and seek help when needed, and never give up on the quest for understanding. With dedication and perseverance, you can master the art of solving differential equations and unlock a world of possibilities.