Chapter 9: First-Order Differential Equations - Practice Exercises - Page 557: 2

$ \ln |y|=\dfrac{e^{x^2}}{2} +c$

Re-arrange the given equation as: $y'=(xy)(e^{x^2})$ and $\int \dfrac{1}{y}dy=\int (xe^{x^2}) dx$ or,$\int \dfrac{1}{y}dy=\dfrac{1}{2} \int (2x) (e^{x^2}) dx$ Consider $x^2=a$ or, $da=(2x) dx$ $\int \dfrac{1}{y}dy=\dfrac{1}{2} \int (2x) (e^{x^2}) dx$ or, $ \ln |y|=\dfrac{e^{a}}{2} +c$ So, $ \ln |y|=\dfrac{e^{x^2}}{2} +c$