We are given the initial-value problem:

$(\ln x) \frac{dy}{dx} + \frac{1}{x} y = 3x^3$ with the condition $y(e) = 7$, and $x > 1$.

Step 1: Rewrite the equation

This is a first-order linear differential equation of the form: $\frac{dy}{dx} + P(x)y = Q(x)$ where $P(x) = \frac{1}{x \ln x}$ and $Q(x) = \frac{3x^3}{\ln x}$.

Now, divide the entire equation by $\ln x$ (since $\ln x \neq 0$ for $x > 1$): $\frac{dy}{dx} + \frac{1}{x \ln x} y = \frac{3x^3}{\ln x}$

Step 2: Find the integrating factor

The integrating factor $\mu(x)$ is given by: $\mu(x) = \exp\left( \int \frac{1}{x \ln x} dx \right)$ Let's compute this integral. Set $u = \ln x$, so $du = \frac{1}{x} dx$, and the integral becomes: $\int \frac{1}{x \ln x} dx = \int \frac{1}{u} du = \ln |u| = \ln (\ln x)$ Thus, the integrating factor is: $\mu(x) = \exp(\ln (\ln x)) = \ln x$

Step 3: Multiply through by the integrating factor

Multiplying both sides of the differential equation by $\ln x$: $\ln x \frac{dy}{dx} + \frac{1}{x} y = 3x^3$

Notice that the left-hand side is the derivative of $y \ln x$ (by the product rule): $\frac{d}{dx}(y \ln x) = 3x^3$

Step 4: Integrate both sides

Now integrate both sides with respect to $x$: $y \ln x = \int 3x^3 dx$ $y \ln x = \frac{3x^4}{4} + C$

Step 5: Solve for $y(x)$

Divide by $\ln x$ to isolate $y(x)$: $y(x) = \frac{3x^4}{4 \ln x} + \frac{C}{\ln x}$

Step 6: Use the initial condition

We are given $y(e) = 7$. Substituting $x = e$ and $\ln e = 1$ into the equation: $7 = \frac{3e^4}{4} + C$ Solve for $C$: $7 = \frac{3e^4}{4} + C \quad \Rightarrow \quad C = 7 - \frac{3e^4}{4}$

Final Solution:

The solution to the differential equation is: $y(x) = \frac{3x^4}{4 \ln x} + \frac{7 - \frac{3e^4}{4}}{\ln x}$

Would you like more details or have any questions about the steps?

Here are 5 related questions for practice:

1. Solve a similar equation where $y(e) = 5$ and the differential equation is $(\ln x) \frac{dy}{dx} + \frac{1}{x} y = 2x^3$.
2. What is the general solution to the equation $y' + P(x)y = Q(x)$?
3. How would the solution change if $y(e) = 0$ instead?
4. Can you solve the same differential equation but for $x > 0$?
5. What is the importance of the integrating factor in solving linear differential equations?

Tip: Always check the initial condition after solving the differential equation to ensure the solution fits.