9.12.1.1 Example 1

$$y^{\prime \prime }-4x{y}^{\prime }+(4x^2-2)y=0$$

Hence $Q=-4x$ and $R=(4x^2-2),f\left(x\right)=0$. Eq(3) becomes

$$M^{\prime }-\frac{1}{2}MQ=0$$

Therefore

$$\begin{array}{rl}M& =e^{\frac{1}{2}\int Qdx}\\ & =e^{\frac{1}{2}\int -4xdx}\\ & =e^{-x^2}\end{array}$$

Now we much check that equation (4) is verified with such $M$.

$$\begin{array}{rl}{M}^{\prime }& =-2x{e}^{-x^2}\\ M^{\prime \prime }& =-2e^{-x^2}-2x(-2e^{-x^2})\\ & =-2e^{-x^2}+4x{e}^{-x^2}\end{array}$$

Substituting these in (4) gives

$$\begin{array}{rl}(-2e^{-x^2}+4x{e}^{-x^2})-e^{-x^2}(4x^2-2)& =0\\ -2e^{-x^2}+4x{e}^{-x^2}+2e^{-x^2}-4x^2{e}^{-x^2}& =0\\ 0& =0\end{array}$$

$M$ is satisfied. Therefore the integrating factor is

$$M=e^{-x^2}$$

Eq (2) now becomes

$$\begin{array}{rl}{\left(My\right)}^{\prime \prime }& =0\\ M{y}^{\prime }& =c_1\\ My& =c_{1}x+c_2\\ y& =\frac{c_{1}x+c_2}{M}\\ & =(c_{1}x+c_2)e^{x^2}\end{array}$$

Which is the same answer found using the more general method of $\mu \left(x\right)$ in the above section but this is simpler when it works since it does not involve solving another ode (the adjoint ode) to find an integrating factor.