Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 24

$$\ln y = \sqrt{t^2+C}$$

Given $$ (\ln y) y'-t y =0$$ Then \begin{aligned}(\ln y) \frac{d y}{d t}-t y &=0 \\(\ln y) \frac{d y}{d t}-t y+t y &=t y \\(\ln y) \frac{d y}{d t} &=t y \\ \frac{\ln y}{y} \frac{d y}{d t} &=t\\ \int \frac{\ln y}{y} dy&=\int tdt\\ \frac{1}{2}(\ln y)^2&=\frac{1}{2}t^2+C\\ \ln y&= \sqrt{t^2+C}\\ \end{aligned}