Chapter 9: First-Order Differential Equations - Section 9.2 - First-Order Linear Equations - Exercises 9.2 - Page 537: 19

$y=-\dfrac{e^{x^2}}{(x+1)}+6e^{x^2}$

Here, we have $y'-2xy=\dfrac{e^{x^2}}{(x+1)^2}$ The integrating factor is: $I=e^{\int -(-2x)dx}=e^{-x^2}$ Now, $e^{-x^2}[y'-2xy]=e^{-x^2}[\dfrac{e^{x^2}}{(x+1)^2}]$ This gives: $y=-\dfrac{1}{(x+1)}e^{x^2}+ce^{x^2}$ Next, apply the initial conditions in the above equation, we have $c=6$ Hence, $y=-\dfrac{e^{x^2}}{(x+1)}+6e^{x^2}$