Example 1 \[ y^{\prime \prime }-4xy^{\prime }+\left ( 4x^{2}-2\right ) y=0 \] Hence \(Q=-4x\) and \(R=\left ( 4x^{2}-2\right ) ,f\left ( x\right ) =0\). Eq(3) becomes\[ M^{\prime }-\frac {1}{2}MQ=0 \] Therefore\begin {align*} M & =e^{\frac {1}{2}\int Qdx}\\ & =e^{\frac {1}{2}\int -4xdx}\\ & =e^{-x^{2}} \end {align*}

Now we much check that equation (4) is verified with such \(M\).\begin {align*} M^{\prime } & =-2xe^{-x^{2}}\\ M^{\prime \prime } & =-2e^{-x^{2}}-2x\left ( -2e^{-x^{2}}\right ) \\ & =-2e^{-x^{2}}+4xe^{-x^{2}} \end {align*}

Substituting these in (4) gives\begin {align*} \left ( -2e^{-x^{2}}+4xe^{-x^{2}}\right ) -e^{-x^{2}}\left ( 4x^{2}-2\right ) & =0\\ -2e^{-x^{2}}+4xe^{-x^{2}}+2e^{-x^{2}}-4x^{2}e^{-x^{2}} & =0\\ 0 & =0 \end {align*}

\(M\) is satisfied. Therefore the integrating factor is \[ M=e^{-x^{2}}\] Eq (2) now becomes\begin {align*} \left ( My\right ) ^{\prime \prime } & =0\\ My^{\prime } & =c_{1}\\ My & =c_{1}x+c_{2}\\ y & =\frac {c_{1}x+c_{2}}{M}\\ & =\left ( c_{1}x+c_{2}\right ) e^{x^{2}} \end {align*}

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.