Hey guys! Let's dive into solving a classic differential equation. We're given that y'(t) = 1 + y(t)^2, and our mission, should we choose to accept it, is to find the solution y(t). This type of problem pops up quite often in calculus and differential equations courses, so understanding how to tackle it is super useful. Ready? Let's get started!

Understanding the Differential Equation

First off, let's break down what the differential equation y'(t) = 1 + y(t)^2 actually means. In simple terms, it's telling us that the rate of change of the function y with respect to t (that's what y'(t) signifies) is equal to 1 plus the square of the function y itself. This is a first-order differential equation because it involves only the first derivative of y. Recognizing this form is crucial because it dictates the methods we can use to solve it.

The key here is to realize that this is a separable differential equation. A separable equation is one where we can rearrange the terms so that all the y terms are on one side and all the t terms are on the other. This allows us to integrate both sides independently, which is a major step towards finding our solution. Identifying these types of equations early on can save you a ton of headache and wasted effort. Seriously, it's like finding the golden ticket in a chocolate factory of math problems.

Also, keep in mind that the '+1' in the equation is what makes it non-linear. If we had something like y'(t) = ky(t), that would be a linear equation, solvable with different techniques. The non-linearity here means we need to be a bit more clever in our approach. But don't worry, we've got this! Think of it like leveling up in a game; the challenges get harder, but so does the satisfaction of overcoming them.

Separating Variables

The heart of solving this differential equation lies in separating the variables. This means getting all the y terms on one side and all the t terms on the other. Starting with y'(t) = 1 + y(t)^2, we can rewrite y'(t) as dy/dt. Now our equation looks like dy/dt = 1 + y^2. To separate the variables, we need to get all the y’s on one side and all the t’s on the other. We can do this by dividing both sides by (1 + y^2) and multiplying both sides by dt. This gives us:

dy / (1 + y^2) = dt

See how neat that is? All the y’s are now cozy on the left side, and the lone dt is chilling on the right. This separation is the linchpin of our solution strategy. Without it, we’d be stuck in a maze of derivatives and integrals. It's like sorting your laundry before you wash it – separating the colors prevents a total disaster!

Integrating Both Sides

Now that we've successfully separated our variables, the next step is to integrate both sides of the equation. We have:

∫ dy / (1 + y^2) = ∫ dt

Let's tackle the left side first. The integral of 1 / (1 + y^2) with respect to y is a standard integral that you should recognize. It's the arctangent function, also known as the inverse tangent. So, the integral of dy / (1 + y^2) is arctan(y).

On the right side, we have the integral of dt, which is simply t. But don't forget to add the constant of integration, which we'll call C. So, the integral of dt is t + C. Putting it all together, we have:

arctan(y) = t + C

This equation is a significant milestone. We've successfully integrated both sides and now have a relationship between y and t. But we're not quite done yet. Our goal is to find y(t), so we need to isolate y.

Solving for y(t)

To isolate y in the equation arctan(y) = t + C, we need to get rid of the arctangent function. The inverse of the arctangent function is the tangent function. So, we'll apply the tangent function to both sides of the equation:

tan(arctan(y)) = tan(t + C)

The tangent of the arctangent of y is simply y. So, we have:

y = tan(t + C)

Therefore, the solution to the differential equation y'(t) = 1 + y^2 is:

y(t) = tan(t + C)

This is our general solution. The C represents an arbitrary constant, which means there are infinitely many solutions to this differential equation, each corresponding to a different value of C. To find a specific solution, we would need an initial condition, such as y(0) = some value. This initial condition would allow us to solve for C and nail down a unique solution.

Applying an Initial Condition (Example)

Let's say we're given the initial condition y(0) = 1. This means that when t = 0, y = 1. We can use this information to find the value of C in our general solution y(t) = tan(t + C).

Plug in t = 0 and y = 1:

1 = tan(0 + C)

1 = tan(C)

To find C, we need to find the angle whose tangent is 1. We know that tan(π/4) = 1. Therefore:

C = π/4

Now we can plug this value of C back into our general solution to get the specific solution that satisfies the initial condition y(0) = 1:

y(t) = tan(t + π/4)

So, this is the particular solution to the differential equation with the given initial condition. Remember, the initial condition is like a specific starting point that pins down one particular curve out of the infinite family of solutions.

Verifying the Solution

It's always a good idea to verify our solution to make sure we didn't make any mistakes along the way. To verify that y(t) = tan(t + C) is indeed a solution to y'(t) = 1 + y(t)^2, we need to find y'(t) and plug it into the equation.

First, let's find the derivative of y(t) = tan(t + C) with respect to t. The derivative of tan(u) is sec^2(u) * du/dt. In our case, u = t + C, so du/dt = 1. Therefore:

y'(t) = sec^2(t + C)

Now, we need to show that y'(t) = 1 + y(t)^2. We know that y(t) = tan(t + C), so y(t)^2 = tan^2(t + C). Our equation becomes:

sec^2(t + C) = 1 + tan^2(t + C)

This is a trigonometric identity! Recall that sec^2(x) = 1 + tan^2(x) for any angle x. Therefore, our equation holds true, and we've verified that y(t) = tan(t + C) is indeed a solution to the differential equation y'(t) = 1 + y(t)^2.

Common Mistakes to Avoid

When solving differential equations like this, there are a few common pitfalls to watch out for:

1. Forgetting the Constant of Integration: This is a classic mistake. Always remember to add the constant of integration (C) after performing an indefinite integral. Forgetting it means you're missing an entire family of solutions.
2. Incorrectly Separating Variables: Make sure you correctly separate the variables before integrating. A mistake here can lead to a completely wrong answer.
3. Not Recognizing Standard Integrals: Knowing your standard integrals (like the integral of 1 / (1 + y^2)) is crucial. Make a cheat sheet if you need to!
4. Algebra Errors: Be careful with your algebra. A small mistake can throw off the entire solution.
5. Not Verifying the Solution: Always verify your solution by plugging it back into the original differential equation.

Conclusion

So, to wrap it all up, we've successfully found the solution to the differential equation y'(t) = 1 + y(t)^2. We learned that y(t) = tan(t + C). We explored how to separate variables, integrate both sides, solve for y(t), apply an initial condition, and verify our solution. Remember to watch out for those common mistakes! With practice, these types of problems will become second nature. Keep practicing, and you'll be solving differential equations like a pro in no time. Keep up the great work, and happy solving!