\(\displaystyle (x^{2}-9)y'+xy=2x(x^{2}-9)\)
Divide by \(\displaystyle (x^{2}-9)\):
\(\displaystyle y'=\frac{x}{x^{2}-9}y=2x\)
The integating factor is \(\displaystyle e^{\int \frac{x}{x^{2}-9}dx}=\sqrt{x^{2}-9}\)
Multiplying through by the I.F., we get:
\(\displaystyle \frac{d}{dx}[y\sqrt{x^{2}-9}]=2x\sqrt{x^{2}-9}\)
Integrate. The left side cancels the derivative because we have the anti-derivative of a derivative:
\(\displaystyle y\sqrt{x^{2}-9}=\frac{2}{3}(x^{2}-9)^{\frac{3}{2}}+C\)
Now, solve for y and use the initial condition to find C. Then, you're done.
