- Effective on any linear [[Ordinary Differential Equations (ODEs)|ODE]] - Finds secondary solutions of an ODE given that we found one solution Uses the product of a known solution and a "helper" function. By substituting this product into the differential equation and applying the product rule repeatedly, all terms of the undifferentiated function cancel, leaving behind a lower order equation for the first derivative of the helper function, which can then by solved and integrated to find the exact value of the helper function. # Example Assuming $p$, $q$, and $r$ are given, continuous functions, we have: $ p(t)y''+q(t)y'+r(t)y=0 $ We assume that we have one solution $y_1(t)$. Another solution can be found with the form: $ y_2=u(t)y_1(t) $ Then differentiate and plug into the original differential equation: $ \begin{align} y'_2&=u'y_1+uy'_1\\ y''_2&=u''y_1+2u'y'_1+uy''_1 \end{align} $ $ \begin{align} p(t)y_2''+q(t)y_2'+r(t)y_2&=0\\ p(t)\left[u''y_1+2u'y'_1+uy''_1\right]+q(t)\left[u'y_1+uy'_1\right]+r(t)uy_1&=0 \end{align} $ Now rearrange in terms of derivatives of $u$: $ p(t)y_1u''+2\left[2py'_1+q(t)y_1\right]u'+\left[p(t)y''_1+q(t)y'_1+r(t)y_1\right]u=0 $ Because $y_1$ is a solution to the original differential equation, the coefficient of $u$ is zero, yielding: $ p(t)y_1u''+\left[2py'_1+q(t)y_1\right]u'=0 $ This is now a first order equation in $u'$, and can be written in terms of $u$: $ \begin{align} \frac{u''}{u'}&=-\frac{\left[2py'_1+q(t)y_1\right]}{y_1p(t}\\ \int\frac{u''}{u'}dt&=\int-\frac{\left[2py'_1+q(t)y_1\right]}{y_1p(t}dt\\ \ln(u')&=\int-\frac{\left[2py'_1+q(t)y_1\right]}{y_1p(t}dt\\ u&=\int \exp\left(\int-\frac{\left[2py'_1+q(t)y_1\right]}{y_1p(t}dt\right)\,dt \end{align} $