\( y \ln x \frac { d x } { d y } = ( \frac { y + 1 } { x } ) ^ { 2 } \)

To solve the differential equation \[ y \ln x \frac{dx}{dy} = \left( \frac{y + 1}{x} \right)^2, \] we can start by rearranging the equation to isolate \(\frac{dx}{dy}\): \[ \frac{dx}{dy} = \frac{(y + 1)^2}{x^2 y \ln x}. \] Next, we can separate the variables \(x\) and \(y\). We rewrite the equation as: \[ x^2 y \ln x \, dx = (y + 1)^2 \, dy. \] Now, we can integrate both sides. However, before we proceed with integration, we can simplify the equation further. Let's express it in a more manageable form: \[ \frac{dx}{(y + 1)^2} = \frac{dy}{x^2 y \ln x}. \] Now, we can integrate both sides. The left side can be integrated with respect to \(x\) and the right side with respect to \(y\). However, this equation is quite complex, and it may be beneficial to look for a substitution or a specific solution. Let's try to find a solution by assuming a specific form for \(x\) in terms of \(y\). Assume \(x = e^u\), then \(\ln x = u\) and \(dx = e^u du\). Substituting this into the equation gives: \[ y u e^u \frac{du}{dy} = \left( \frac{y + 1}{e^u} \right)^2. \] This simplifies to: \[ y u e^u \frac{du}{dy} = \frac{(y + 1)^2}{e^{2u}}. \] Multiplying both sides by \(e^{2u}\) gives: \[ y u e^{3u} \frac{du}{dy} = (y + 1)^2. \] Now, we can separate variables again: \[ \frac{du}{(y + 1)^2} = \frac{dy}{y u e^{3u}}. \] This form is still complex, and it may be more effective to look for particular solutions or numerical methods to solve it. If you have specific boundary conditions or initial values, please provide them, and we can look for a more tailored solution. Otherwise, numerical methods or software tools may be required to find a general solution to this equation.