This is 1st order and nonlinear. But it can be turned into a first order linear differential equation which can be solved with an integrating factor. First divide each side by \(y^3\).
\[y^{-3}y'+2y^{-2}=x^2\]Now we can see that we have \(y^{-2}\) and its derivative is proportional to \(y^{-3}y'\) which we also see. So to turn this into first order, we can make the substitution \(u=y^{-2},u'=-2y^{-3}y'\) after multiplying each side by \(-2\).
\[-2y^{-3}y'-4y^{-2}=-2x^2\] \[u'-4u=-2x^2\]The integrating factor would be \(e^{\int-4dx}=e^{-4x}\). Multiply each side by this.
\[e^{-4x}u'-4e^{-4x}u=-2x^2e^{-4x}\]Now we can see that the left side is in the form of the product rule, which we can reverse.
\[(e^{-4x}u)'=-2x^2e^{-4x}\]Next, we integrate both sides so we need to use integration by parts.
\[e^{-4x}u=-\int2x^2e^{-4x}dx=\int\left(\frac{-1}{4}e^{-4x}\cdot4x\right)dx -(2x^2)\left(\frac{-1}{4}e^{-4x}\right)\] \[=-\int xe^{-4x}dx+\frac{1}{2}x^2e^{-4x} =\int\left(\frac{-1}{4}e^{-4x}\right)dx-x\left(\frac{-1}{4}e^{-4x}\right) +\frac{1}{2}x^2e^{-4x}\] \[=\frac{1}{16}e^{-4x}+\frac{x}{4}e^{-4x}+\frac{1}{2}x^2e^{-4x}+C =\frac{e^{-4x}}{16}\left(1+4x+8x^2\right)+C\]Now multiply each side by \(e^{4x}\) and substitute \(u=y^{-2}\) so we have.
\[y^{-2}=\frac{1+4x+8x^2+16Ce^{4x}}{16}\]Then we flip this, take the square root, and replace \(16C\) with some constant \(c_1\) so the solution is.
\[y=\frac{\pm4}{\sqrt{1+4x+8x^2+c_1e^{4x}}}\]