To solve the given differential equation:
[tex]\[ y \, dx + x \ln(x) \, dy = 0, \][/tex]
we start by expressing it in a more standard form for solving first-order differential equations.
First, rearrange the terms to make it easier to work with:
[tex]\[ x \ln(x) \, dy = -y \, dx. \][/tex]
Next, divide both sides by [tex]\(x \ln(x) y\)[/tex] to separate the variables [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ \frac{dy}{y} = -\frac{dx}{x \ln(x)}. \][/tex]
Now, we integrate both sides. To integrate the left side, we have:
[tex]\[ \int \frac{1}{y} \, dy. \][/tex]
This integral is:
[tex]\[ \ln|y| + C_1. \][/tex]
For the right side, we need to integrate:
[tex]\[ \int -\frac{1}{x \ln(x)} \, dx. \][/tex]
This integral can be solved using the substitution [tex]\( u = \ln(x) \)[/tex], hence [tex]\( du = \frac{1}{x} \, dx \)[/tex]. The integral becomes:
[tex]\[ -\int \frac{1}{u} \, du, \][/tex]
which simplifies to:
[tex]\[ -\ln|u|. \][/tex]
Since [tex]\( u = \ln(x) \)[/tex], this becomes:
[tex]\[ -\ln|\ln(x)| + C_2. \][/tex]
Equating both integrals and combining the constants:
[tex]\[ \ln|y| = -\ln|\ln(x)| + C. \][/tex]
We can exponentiate both sides to remove the logarithms:
[tex]\[ |y| = e^{-\ln|\ln(x)| + C} = e^C \cdot e^{-\ln|\ln(x)|}. \][/tex]
Let [tex]\( e^C = C_1 \)[/tex], a new constant of integration, and simplify:
[tex]\[ |y| = \frac{C_1}{\ln(x)}. \][/tex]
Thus, the general solution to the differential equation is:
[tex]\[ y(x) = \frac{C_1}{\ln(x)}. \][/tex]
Therefore, we write this as:
[tex]\[ y = \frac{C_1}{\ln(x)}, \][/tex]
which provides the general solution to the given differential equation.
