All second order linear ODEs have symmetries of the form [xi=0, eta=F(x)]. Actually, F(x) is always a solution of the related homogeneous ODE. There is no general scheme for determining F(x); see dsolve,linear).

When a symmetry of the form [xi=0, eta=F(x)] is found, this information is enough to integrate the homogeneous ODE (see Murphy's book, p. 88).

In the case of nonhomogeneous ODEs, you can do the following:

1) look for F(x) as a symmetry of the homogeneous ODE;

2) solve the homogeneous ODE using this information;

3) set each of _C1 and _C2 equal to 0 and 1 in the answer of the previous step, in order to obtain the two linearly independent solutions of the homogeneous ODE;

4) use these two independent solutions of the homogeneous ODE to build the general solution to the nonhomogeneous ODE (see Bluman and Kumei, Symmetries and Differential Equations, p. 132 and ?dsolve,references).

$\mathrm{with}\left(\mathrm{DEtools}\,\mathrm{odeadvisor}\,\mathrm{symgen}\right)$

$\left[\mathrm{odeadvisor}\,\mathrm{symgen}\right]$

$\mathrm{ode}\left[1\right]≔\mathrm{diff}\left(y\left(x\right)\,x\,x\right)=a\mathrm{diff}\left(y\left(x\right)\,x\right)-\frac{\frac{1}{\ln \left(x\right)}y\left(x\right)}{x^2}-\frac{\frac{a}{\ln \left(x\right)}y\left(x\right)}{x}$

${\mathrm{ode}}_1≔\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=a\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-\frac{y\left(x\right)}{\ln \left(x\right)x^2}-\frac{ay\left(x\right)}{\ln \left(x\right)x}$

$\mathrm{odeadvisor}\left(\mathrm{ode}\left[1\right]\right)$

$\left[\left[\mathrm{\_2nd\_order}\,\mathrm{\_with\_linear\_symmetries}\right]\,\left[\mathrm{\_2nd\_order}\,\mathrm{\_linear}\,\mathrm{\_with\_symmetry\_[0,F(x)]}\right]\right]$

$\mathrm{dsolve}\left(\mathrm{ode}\left[1\right]\right)$

$y\left(x\right)=\left(\int \frac{{\ⅇ}^{ax}}{{\ln \left(x\right)}^2}\ⅆx\mathrm{c\_\_1}+\mathrm{c\_\_2}\right)\ln \left(x\right)$

$\mathrm{ode}\left[2\right]≔\mathrm{diff}\left(y\left(x\right)\,x\,x\right)=\frac{\mathrm{diff}\left(F\left(x\right)\,x\,x\right)y\left(x\right)}{F\left(x\right)}+H\left(x\right)$

${\mathrm{ode}}_2≔\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\frac{\left(\frac{{\ⅆ}^2}{\ⅆx^2}F\left(x\right)\right)y\left(x\right)}{F\left(x\right)}+H\left(x\right)$

$\mathrm{dsolve}\left(\mathrm{ode}\left[2\right]\,y\left(x\right)\right)$

$y\left(x\right)=\int \frac{1}{{F\left(x\right)}^2}\ⅆxF\left(x\right)\mathrm{c\_\_2}+F\left(x\right)\mathrm{c\_\_1}+F\left(x\right)\left(\int \frac{1}{{F\left(x\right)}^2}\ⅆx\int H\left(x\right)F\left(x\right)\ⅆx-\int \int \frac{1}{{F\left(x\right)}^2}\ⅆxF\left(x\right)H\left(x\right)\ⅆx\right)$

$\mathrm{ode}\left[3\right]≔\mathrm{diff}\left(y\left(x\right)\,\mathrm{`\$`}\left(x\,2\right)\right)-y\left(x\right)=F\left(x\right)$

${\mathrm{ode}}_3≔\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-y\left(x\right)=F\left(x\right)$

$\mathrm{homogeneous\_ode}≔\mathrm{diff}\left(y\left(x\right)\,\mathrm{`\$`}\left(x\,2\right)\right)-y\left(x\right)$

$\mathrm{homogeneous\_ode}≔\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)-y\left(x\right)$

$\mathrm{ans\_h}≔\mathrm{dsolve}\left(\mathrm{homogeneous\_ode}\right)$

$\mathrm{ans\_h}≔y\left(x\right)=\mathrm{c\_\_1}{\ⅇ}^{-x}+\mathrm{c\_\_2}{\ⅇ}^x$

$\mathrm{sol\_1}≔\mathrm{rhs}\left(\mathrm{subs}\left(\left[\mathrm{\_C1}=1\,\mathrm{\_C2}=0\right]\,\mathrm{ans\_h}\right)\right)$

$\mathrm{sol\_1}≔{\ⅇ}^{-x}$

$\mathrm{sol\_2}≔\mathrm{rhs}\left(\mathrm{subs}\left(\left[\mathrm{\_C1}=0\,\mathrm{\_C2}=1\right]\,\mathrm{ans\_h}\right)\right)$

$\mathrm{sol\_2}≔{\ⅇ}^x$

$\mathrm{ANS}≔\left(\mathrm{s1}\,\mathrm{s2}\,F\right)\mapsto y\left(x\right)=\mathrm{\_C1}\cdot \mathrm{s2}+\mathrm{\_C2}\cdot \mathrm{s1}+\mathrm{s2}\cdot \mathrm{int}\left(\frac{F\cdot \mathrm{s1}}{W\left(\mathrm{s1}\,\mathrm{s2}\right)}\,x\right)-\mathrm{s1}\cdot \mathrm{int}\left(\frac{F\cdot \mathrm{s2}}{W\left(\mathrm{s1}\,\mathrm{s2}\right)}\,x\right)$

$\mathrm{ANS}≔\left(\mathrm{s1}\,\mathrm{s2}\,F\right)\mapsto y\left(x\right)=\mathrm{c\_\_1}\cdot \mathrm{s2}+\mathrm{c\_\_2}\cdot \mathrm{s1}+\mathrm{s2}\cdot \int \frac{F\cdot \mathrm{s1}}{W\left(\mathrm{s1}\,\mathrm{s2}\right)}\ⅆx-\mathrm{s1}\cdot \int \frac{F\cdot \mathrm{s2}}{W\left(\mathrm{s1}\,\mathrm{s2}\right)}\ⅆx$

$W≔\left(\mathrm{s1}\,\mathrm{s2}\right)\→\mathrm{simplify}\left(\mathrm{s1}\left(\frac{\partial }{\partial x}\mathrm{s2}\right)-\mathrm{s2}\left(\frac{\partial }{\partial x}\mathrm{s1}\right)\right)$

$W≔\left(\mathrm{s1}\,\mathrm{s2}\right)\→\mathrm{simplify}\left(\mathrm{s1}\left(\frac{\partial }{\partial x}\mathrm{s2}\right)-\mathrm{s2}\left(\frac{\partial }{\partial x}\mathrm{s1}\right)\right)$

$\mathrm{ans}≔\mathrm{ANS}\left(\mathrm{sol\_1}\,\mathrm{sol\_2}\,F\left(x\right)\right)$

$\mathrm{ans}≔y\left(x\right)=\mathrm{c\_\_1}{\ⅇ}^x+\mathrm{c\_\_2}{\ⅇ}^{-x}+{\ⅇ}^x\int \frac{F\left(x\right){\ⅇ}^{-x}}{2}\ⅆx-{\ⅇ}^{-x}\int \frac{F\left(x\right){\ⅇ}^x}{2}\ⅆx$

$\mathrm{odetest}\left(\mathrm{ans}\,\mathrm{ode}\left[3\right]\right)$

$0$