To solve first-order differential equations using integration, we follow a set of easy steps. These steps help us change the tricky differential equation into a form we can manage. This is especially important when the equation can be separated or easily integrated.

Step 1: Analyze the Equation

First, we figure out what kind of first-order differential equation we have.

The two common types are:

1. Separable equations
2. Linear equations

Knowing the type is important because the way we solve them can vary a lot.

A first-order differential equation usually looks like this:

$$\frac{dy}{dx} = f(x, y)$$

or:

$$F(x, y, \frac{dy}{dx}) = 0.$$

For separable equations, we can express the function $f(x, y)$ as a product of a function of $x$ and a function of $y$:

$$\frac{dy}{dx} = g(x)h(y).$$

If we can’t separate the equation like this, we’ll need to use other methods such as integrating factors for linear equations or substitution methods.

Step 2: Separate Variables (for Separable Equations)

Once we confirm that our differential equation is separable, we need to separate the variables.

This usually means rearranging the equation to put the $y$ terms on one side and the $x$ terms on the other:

$$\frac{dy}{h(y)} = g(x) \, dx.$$

This step is important because it prepares us to integrate each side individually in the next step.

Step 3: Integrate Both Sides

Now, we integrate both sides of the equation.

For our separated equation, we write it like this:

$$\int \frac{1}{h(y)} \, dy = \int g(x) \, dx.$$

This step may involve different methods of integration, depending on how complex the functions are. After integrating, we end up with expressions for both sides that include a constant, usually represented by $C$.

Step 4: Solve for y (if necessary)

After we finish integrating, we might still have $y$ described in terms of $x$.

Sometimes, we can solve for $y$ directly. This means rearranging the equation to get $y$ alone on one side. The result should give us a function of $x$, which shows the general solution of the differential equation.

If we can’t easily isolate $y$, it’s okay to leave it in implicit form. Both ways can be correct, depending on what the problem requires.

Step 5: Apply Initial Conditions (if provided)

Sometimes we have initial conditions, which tell us a specific value of $y$ when $x$ equals a certain number.

If we have this information, we plug those values into our general solution to solve for $C$. This gives us a specific solution that corresponds to the initial condition we were given.

Step 6: Validate the Solution

Once we have our solution, we should check that it’s correct.

We can do this by taking the derivative of our resulting function. If the left-hand side of the original equation equals the right-hand side after substituting back in, we can be sure our solution is correct.

Example of a Separable First-Order Differential Equation

Let’s look at a simple example to see how these steps work:

$$\frac{dy}{dx} = y \sin(x).$$

1. Analyze the Equation: This is a separable equation, with $f(x, y) = y \sin(x)$.

2. Separate Variables: We rearrange it to:

   $$\frac{1}{y} \, dy = \sin(x) \, dx.$$

3. Integrate Both Sides: Now we integrate both sides:

   $$\int \frac{1}{y} \, dy = \int \sin(x) \, dx.$$

   This gives us:

   $$\ln |y| = -\cos(x) + C.$$

4. Solve for $y$: By removing the logarithm, we get:

   $$|y| = e^{-\cos(x) + C}.$$

   We can deal with the absolute value based on the sign of $C$.

5. Apply Initial Conditions: If we had an initial condition like $y(0) = 1$, we substitute this in:

   $$1 = e^{-\cos(0) + C} = e^{-1 + C} \implies C = 1.$$

6. Validate the Solution: Finally, we differentiate:

   $$\frac{dy}{dx} = y \sin(x),$$

   which confirms that our solution works.

Conclusion

To sum it up, solving first-order differential equations with integration means following clear steps: analyzing, separating variables, integrating, solving for $y$, applying initial conditions, and validating the solution. Learning these steps helps students and professionals handle different problems in advanced calculus and differential equations, linking integration methods to real-world applications. It is essential for students in calculus classes to understand these steps, as they form a crucial part of solving math problems.