Solve the differential equation \[ \frac{d x}{d t}=\frac{x \ln x}{t} \] Assume \( x, t>0 \), and use the initial condition \( x(1)=5 \) \( x= \)

To solve the differential equation \[ \frac{d x}{d t} = \frac{x \ln x}{t}, \] we can use the method of separation of variables. First, rearranging gives us: \[ \frac{dx}{x \ln x} = \frac{dt}{t}. \] Next, we integrate both sides. The left side can be integrated using the substitution \( u = \ln x \), which means \( du = \frac{1}{x} dx \), and thus, \( dx = x du = e^u du \). So, \[ \int \frac{dx}{x \ln x} = \int \frac{1}{\ln x} d(\ln x) = \int \frac{du}{u} = \ln |\ln x| + C_1. \] For the right side, we have: \[ \int \frac{dt}{t} = \ln |t| + C_2. \] Setting \( C = C_2 - C_1 \), we combine the integrals to get: \[ \ln |\ln x| = \ln |t| + C. \] Exponentiating both sides yields: \[ |\ln x| = K |t| e^C, \] where \( K = e^C \) is a positive constant. Assuming \( x > 0 \) means \( |\ln x| = \ln x \), thus: \[ \ln x = Kt. \] Exponentiating gives: \[ x = e^{Kt} = e^{x_0} t^{K}, \] where \( x_0 \) is derived from the constant terms. To use the initial condition \( x(1) = 5 \): \[ 5 = e^{K \cdot 1} \implies K = \ln 5. \] So we can express \( x(t) \) in terms of \( K \): \[ x(t) = 5 t^{\ln 5} \text{ or } x(t) = 5t^{\ln 5}. \] Therefore, the solution to the differential equation with the given initial condition is: \[ \boxed{x = 5 t^{\ln 5}}. \]