We are given the initial value problem (IVP):

$x y' = \ln x, \quad y(1) = -2$

Step 1: Rewrite as a Differential Equation

Rewriting the given equation in standard form:

$y' = \frac{\ln x}{x}$

This is a separable differential equation.

Step 2: Integrate Both Sides

Rewriting:

$\frac{dy}{dx} = \frac{\ln x}{x}$

Separating variables:

$dy = \frac{\ln x}{x} dx$

Now, integrate both sides:

$\int dy = \int \frac{\ln x}{x} dx$

Using the substitution $u = \ln x$, so that $du = \frac{1}{x} dx$, we rewrite the integral:

$\int u \, du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C$

Thus, we obtain:

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

Step 3: Apply the Initial Condition

Using $y(1) = -2$, we substitute $x = 1$:

$-2 = \frac{(\ln 1)^2}{2} + C$

Since $\ln 1 = 0$, this simplifies to:

$-2 = 0 + C$

Thus, $C = -2$.

Step 4: Write the Explicit Solution

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

Final Answer:

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

Would you like a detailed explanation of any step?

Related Problems:

1. Solve $x y' = e^x$ with $y(1) = 0$.
2. Find the general solution to $y' = \frac{x^2}{y}$.
3. Solve $y' + 2y = x^2$ using an integrating factor.
4. Find the solution to $y' = e^x y$ with $y(0) = 1$.
5. Solve $xy' + y = \sin x$ using the method of integrating factors.

Tip:

For separable differential equations, always check if variables can be neatly separated before choosing a different method!