We are given the following initial value problem (IVP):

$\frac{dy}{dx} = y e^{3x}, \quad y(0) = 10e$

Step 1: Solve the differential equation

This is a separable differential equation, so we can rewrite it as:

$\frac{1}{y} \, dy = e^{3x} \, dx$

Now integrate both sides:

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

The left-hand side gives:

$\ln|y| = \int e^{3x} \, dx$

The right-hand side integrates to:

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

Where $C$ is the constant of integration.

Step 2: Solve for $y$

Exponentiate both sides to get rid of the logarithm:

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

Since $e^C$ is just a constant, we can replace it with a new constant $A$, giving:

$y = A e^{\frac{1}{3}e^{3x}}$

Step 3: Apply the initial condition

We are given that $y(0) = 10e$. Substituting $x = 0$ into the solution:

$10e = A e^{\frac{1}{3}e^0} = A e^{\frac{1}{3}}$

Thus, we can solve for $A$:

$A = \frac{10e}{e^{\frac{1}{3}}} = 10e^{1 - \frac{1}{3}} = 10e^{\frac{2}{3}}$

Step 4: Write the final solution

Substitute this value of $A$ back into the equation for $y$:

$y = 10e^{\frac{2}{3}} e^{\frac{1}{3}e^{3x}}$

Thus, the explicit particular solution is:

$y = 10e^{\frac{2}{3} + \frac{1}{3}e^{3x}}$

This is the explicit solution for the given initial value problem.

Would you like further explanation of any step or more details?

Related questions:

1. What is the general process for solving separable differential equations?
2. How does the initial condition affect the solution of a differential equation?
3. Why do we need to apply logarithmic rules when solving differential equations?
4. How would the solution change if the initial condition were different?
5. Can you solve non-separable differential equations using this method?

Tip:

Always double-check the constants of integration when solving initial value problems to ensure you properly account for the given conditions.