ode internal name "linear_second_order_ode_solved_by_an_M_integrating_factor"
This is another method to find integrating factor method for the second order ode. This method of finding an integrating factor is not a general one like the above using \(\mu \left ( x\right ) \) but it is easier to check. This is tried first and if this does not work, then the above will be tried.
Given the ode, normalized so that the coefficient of \(y^{\prime \prime }\) is one\begin {equation} y^{\prime \prime }+Q\left ( x\right ) y^{\prime }+R\left ( x\right ) y=f\left ( x\right ) \tag {1} \end {equation} Let there exists an integrating factor \(M\left ( x\right ) \) such that\begin {equation} \left ( M\left ( x\right ) y\right ) ^{\prime \prime }=M\left ( x\right ) f\left ( x\right ) \tag {2} \end {equation} Then it can be integrated twice and solved. To find \(M\), the above becomes\begin {align} \left ( M^{\prime }y+My^{\prime }\right ) ^{\prime } & =Mf\nonumber \\ M^{\prime \prime }y+M^{\prime }y^{\prime }+M^{\prime }y^{\prime }+My^{\prime \prime } & =Mf\nonumber \\ My^{\prime \prime }+y^{\prime }\left ( 2M^{\prime }\right ) +M^{\prime \prime }y & =Mf\nonumber \\ y^{\prime \prime }+y^{\prime }\left ( 2\frac {M^{\prime }}{M}\right ) +\frac {M^{\prime \prime }}{M}y & =f \tag {2A} \end {align}
Comparing (2A) to (1) gives\begin {align*} 2\frac {M^{\prime }}{M} & =Q\\ \frac {M^{\prime \prime }}{M} & =R \end {align*}
Or\begin {align} M^{\prime }-\frac {1}{2}MQ & =0\tag {3}\\ M^{\prime \prime }-MR & =0 \tag {4} \end {align}
Starting with (3) gives \(M=e^{\frac {1}{2}\int Qdx}\). If this also satisfies (4), then \(M\) is found by integration. If not, then this method did not work.