Answer
Answer given below.
Work Step by Step
Take the derivatives of the function.
$y(x)=c_1x^{\frac{1}{2}}+3x^2$
$y'(x)=\frac{1}{2}c_1x^{\frac{-1}{2}}+6x$
$y''(x)=\frac{-1}{4}c_1x^{\frac{-3}{2}}+6$
Substituting these functions into the differential equation yields
$\implies 2x^2y''-xy'+y=9x^2$
$\implies 2x^2(\frac{-1}{4}c_1x^{\frac{-3}{2}}+6)-x(\frac{1}{2}c_1x^{\frac{-1}{2}}+6x)+c_1x^{\frac{1}{2}}+3x^2=9x^2$
$\implies \frac{-1}{2}c_1x^{\frac{1}{2}}+12x^2-\frac{1}{2}c_1x^{\frac{1}{2}}-6x^2+c_1x^{\frac{1}{2}}+3x^2-9x^2=0$
$\implies 0=0$
This means that the solution is valid for all real values. The interval linked to this solution is $(−∞,∞)$.
