We are given the following differential equation:

$(\ln y) y' = -x^2 y, \quad y(0) = e.$

Step 1: Rearranging the equation

We first rewrite the given equation in a more manageable form:

$(\ln y) \frac{dy}{dx} = -x^2 y.$

To solve this separable differential equation, we separate variables and integrate.

Step 2: Separation of variables

First, divide both sides by $y$:

$\frac{\ln y}{y} \frac{dy}{dx} = -x^2.$

Rearranging to separate $y$ and $x$:

$\frac{\ln y}{y} dy = -x^2 dx.$

Step 3: Integration

We now integrate both sides.

• The left-hand side requires the integration of $\frac{\ln y}{y}$, which can be done by substitution. Let $u = \ln y$, so $du = \frac{1}{y} dy$. The integral becomes:

$\int \frac{\ln y}{y} dy = \int u \, du = \frac{u^2}{2} = \frac{(\ln y)^2}{2}.$

• The right-hand side is straightforward:

$\int -x^2 dx = -\frac{x^3}{3}.$

Thus, after integrating both sides, we get:

$\frac{(\ln y)^2}{2} = -\frac{x^3}{3} + C.$

Step 4: Solve for $y$

To solve for $y$, multiply both sides by 2:

$(\ln y)^2 = -\frac{2x^3}{3} + 2C.$

Let $2C = C_1$ (since it's an arbitrary constant, it can be renamed). So we have:

$(\ln y)^2 = -\frac{2x^3}{3} + C_1.$

Step 5: Apply the initial condition $y(0) = e$

Substitute $x = 0$ and $y = e$ into the equation to find $C_1$:

$(\ln e)^2 = 0 = -\frac{2(0)^3}{3} + C_1.$

So, $C_1 = 0$. Therefore, the solution simplifies to:

$(\ln y)^2 = -\frac{2x^3}{3}.$

Step 6: Final solution and matching with options

Thus, the final form of the solution is:

$\frac{(\ln y)^2}{2} = -\frac{x^3}{3} + \frac{1}{2}.$

Comparing with the given options, the correct choice is:

Option (b): $\frac{(\ln y)^2}{2} = -\frac{x^3}{3} + \frac{1}{2}.$

Final answer:

The correct answer is (b).

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Would you like more details or have any further questions? Here are some related problems to deepen your understanding:

1. Solve $y' = x^2y^2$ using separation of variables.
2. Solve the separable DE $y' = y/x$, $y(1) = 2$.
3. Find the general solution of $y' = y \ln y$.
4. Solve $x^2y' = y + x^3$, with $y(1) = 0$.
5. Verify the uniqueness of the solution for the initial condition $y(0) = e^2$ in this problem.

Tip: When solving separable differential equations, always ensure that you handle constants carefully during integration, and don’t forget to apply initial conditions if given!