We are tasked with solving the differential equation:

$\frac{dy}{dx} = \frac{x^2 - 1}{e^{3y + 3}}$

Given the initial condition: $y(3) = -1$.

Step 1: Separation of Variables

First, rewrite the differential equation to separate the variables $y$ and $x$:

$e^{3y + 3} \, dy = (x^2 - 1) \, dx$

Step 2: Integration

Integrating both sides of the equation:

$\int e^{3y + 3} \, dy = \int (x^2 - 1) \, dx$

The integral of the left side is:

$\int e^{3y + 3} \, dy = \frac{1}{3} e^{3y + 3}$

The integral of the right side is:

$\int (x^2 - 1) \, dx = \frac{x^3}{3} - x$

Thus, the equation becomes:

$\frac{1}{3} e^{3y + 3} = \frac{x^3}{3} - x + C$

Multiplying through by 3 to simplify:

$e^{3y + 3} = x^3 - 3x + 3C$

Step 3: Solving for the Particular Solution

We now use the initial condition $y(3) = -1$ to solve for the constant $C$.

Substitute $x = 3$ and $y = -1$ into the equation:

$e^{3(-1) + 3} = 3^3 - 3(3) + 3C$

This simplifies to:

$e^{0} = 27 - 9 + 3C$

Since $e^0 = 1$, we get:

$1 = 18 + 3C$

Solving for $C$:

$3C = 1 - 18 = -17 \quad \Rightarrow \quad C = -\frac{17}{3}$

Step 4: Final Solution

Substitute $C = -\frac{17}{3}$ back into the general solution:

$e^{3y + 3} = x^3 - 3x - 17$

Finally, solve for $y$:

$e^{3y + 3} = x^3 - 3x - 17$

Taking the natural logarithm of both sides:

$3y + 3 = \ln(x^3 - 3x - 17)$

Thus, solving for $y$:

$y = \frac{1}{3} \ln(x^3 - 3x - 17) - 1$

This is the particular solution.

Do you have any questions or want more details? Here are some related questions to practice:

1. How can you verify if the solution is correct using the initial condition?
2. What is the general solution to this differential equation before applying the initial condition?
3. What is the process for solving separable differential equations?
4. How do you apply the method of integrating factors in solving differential equations?
5. How does the exponential function $e^{3y + 3}$ behave in a differential equation?

Tip: Always check if the solution satisfies the given initial condition by substituting back into the original equation.