Solving Separable Differential Equation Dx/dt = X^2 + 1/25 With X(0) = -6

Hey there, math enthusiasts! Today, we're diving into the fascinating world of differential equations, specifically tackling a separable differential equation. We're going to break down the process step-by-step, ensuring you not only understand how to solve it but also why each step is crucial. So, grab your thinking caps, and let's get started!

Understanding Separable Differential Equations

Before we jump into the solution, let's quickly recap what separable differential equations are all about. In essence, these equations are those where we can isolate the variables on opposite sides of the equation. Think of it like sorting socks – you want all the left socks on one side and all the right socks on the other. In our case, we want all the 'x' terms on one side and all the 't' terms on the other. This separation is the key to solving these types of equations, and this process allows us to integrate each side independently, ultimately leading us to the general solution. Spotting a separable equation is the first hurdle, and once you've mastered that, you're well on your way to conquering differential equations.

Now, let's take a closer look at the given differential equation: $\frac{dx}{dt} = x^2 + \frac{1}{25}$. Can we separate the variables here? Absolutely! Notice that we have 'x' terms on the right side and 'dx/dt' on the left. Our goal is to get all 'x' terms with 'dx' and all 't' terms with 'dt'. This involves a bit of algebraic manipulation, but it's a straightforward process. By separating the variables, we set the stage for integration, which is the next big step in solving our equation. So, let's move on to the separation process and see how it unfolds.

Separating the Variables

The given differential equation is $\frac{dx}{dt} = x^2 + \frac{1}{25}$. To separate the variables, we need to get all the 'x' terms on one side and all the 't' terms on the other. The first step is to treat $\frac{dx}{dt}$ as a fraction and multiply both sides by 'dt'. This gives us $dx = (x^2 + \frac{1}{25}) dt$. Now, we want to isolate the 'x' terms on the left side. To do this, we divide both sides by $(x^2 + \frac{1}{25})$. This results in $\frac{dx}{x^2 + \frac{1}{25}} = dt$. See? We've successfully separated the variables! All the 'x' terms are now on the left side with 'dx', and all the 't' terms (just 'dt' in this case) are on the right side. This separation is crucial because it allows us to integrate both sides independently, which is the next step in finding the solution. This algebraic manipulation might seem simple, but it's a fundamental technique in solving separable differential equations. Without this separation, we wouldn't be able to proceed with integration. So, make sure you're comfortable with this step before moving on. Remember, practice makes perfect, so try separating variables in other similar equations to solidify your understanding. Now that we've successfully separated the variables, let's move on to the exciting part: integration!

Integrating Both Sides

Alright, guys, we've separated the variables, which means we're ready for the main event: integration! We've got $\frac{dx}{x^2 + \frac{1}{25}} = dt$. Now, we need to integrate both sides of the equation. Let's start with the left side: $\int \frac{dx}{x^2 + \frac{1}{25}}$. This integral might look a bit intimidating at first, but don't worry, we've got this! Recognize that this integral is in the form of $\int \frac{du}{u^2 + a^2}$, where 'u' is our 'x' and 'a' is $\frac{1}{5}$ (since $\frac{1}{25}$ is $(\frac{1}{5})^2$). This is a standard integral that we can solve using the arctangent (tan⁻¹) function. The integral of $\frac{du}{u^2 + a^2}$ is $\frac{1}{a} \arctan(\frac{u}{a}) + C$. Applying this to our integral, we get:

$\int \frac{dx}{x^2 + \frac{1}{25}} = \frac{1}{\frac{1}{5}} \arctan(\frac{x}{\frac{1}{5}}) + C_1 = 5 \arctan(5x) + C_1$

Now, let's tackle the right side: $\int dt$. This is a much simpler integral. The integral of 'dt' with respect to 't' is simply 't'. So, $\int dt = t + C_2$. Now we have integrated both sides, resulting in:

$5 \arctan(5x) + C_1 = t + C_2$

We can combine the constants of integration (C₁ and C₂) into a single constant, let's call it 'C'. So, our equation becomes:

$5 \arctan(5x) = t + C$

This equation represents the general solution to our differential equation. We're not quite done yet, though! We still need to isolate 'x' to get the explicit solution, and then we'll use the initial condition to find the particular solution. So, let's keep pushing forward!

Finding the General Solution for x(t)

Okay, we've reached a crucial point in our journey! We've integrated both sides of the equation and arrived at $5 \arctan(5x) = t + C$. Now, our mission is to isolate 'x' and express it as a function of 't', giving us the general solution x(t). To do this, we need to carefully unwind the equation step by step. First, let's divide both sides by 5:

$\arctan(5x) = \frac{t}{5} + \frac{C}{5}$

Since C is an arbitrary constant, C/5 is also an arbitrary constant. We can rename it to a new constant, let's say 'K', where $K = \frac{C}{5}$. So, the equation becomes:

$\arctan(5x) = \frac{t}{5} + K$

Next, we need to get rid of the arctangent function. To do this, we take the tangent of both sides. Remember, the tangent function is the inverse of the arctangent function, so it will help us isolate the term inside the arctangent:

$\tan(\arctan(5x)) = \tan(\frac{t}{5} + K)$

This simplifies to:

$5x = \tan(\frac{t}{5} + K)$

Now, we're almost there! To finally isolate 'x', we divide both sides by 5:

$x = \frac{1}{5} \tan(\frac{t}{5} + K)$

This is the general solution for x(t). It represents a family of solutions, each corresponding to a different value of the constant K. But remember, we're not just looking for any solution; we're looking for the particular solution that satisfies a specific initial condition. That's our next challenge, and it will allow us to pinpoint the exact solution we need. So, let's move on to the final stage: finding the particular solution!

Applying the Initial Condition x(0) = -6

Alright, we've reached the final stretch! We've got the general solution: $x(t) = \frac{1}{5} \tan(\frac{t}{5} + K)$. Now, we need to find the particular solution that satisfies the initial condition $x(0) = -6$. This means when t = 0, x = -6. Let's plug these values into our general solution:

$-6 = \frac{1}{5} \tan(\frac{0}{5} + K)$

This simplifies to:

$-6 = \frac{1}{5} \tan(K)$

Now, we need to solve for K. Multiply both sides by 5:

$-30 = \tan(K)$

To find K, we take the arctangent (tan⁻¹) of both sides:

$K = \arctan(-30)$

Now that we have the value of K, we can plug it back into our general solution to get the particular solution:

$x(t) = \frac{1}{5} \tan(\frac{t}{5} + \arctan(-30))$

And there you have it! This is the particular solution to the differential equation that satisfies the initial condition x(0) = -6. We've successfully navigated the entire process, from separating variables to integrating and finally applying the initial condition. This journey through differential equations might seem challenging, but with practice and a clear understanding of each step, you can conquer any separable differential equation that comes your way. Keep practicing, and you'll become a master of differential equations in no time!

$x(t) = \frac{1}{5} \tan(\frac{t}{5} + \arctan(-30))$

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