The liesol routine attempts to find a solution to the equation by using Lie methods. See dsolve,Lie.

The first argument is a differential equation in diff or D form and the second argument is the function in the differential equation.

This function is part of the DEtools package, and so it can be used in the form liesol(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[liesol](..).

$\mathrm{with}\left(\mathrm{DEtools}\right)\:$

$\mathrm{ode}≔9{\mathrm{diff}\left(y\left(x\right)\,x\,x\right)}^2\mathrm{diff}\left(y\left(x\right)\,x\,x\,x\,x\,x\right)-45\mathrm{diff}\left(y\left(x\right)\,x\,x\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\,x\right)\mathrm{diff}\left(y\left(x\right)\,x\,x\,x\,x\right)+40\mathrm{diff}\left(y\left(x\right)\,x\,x\,x\right)$

$\mathrm{ode}≔9{\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)}^2\left(\frac{{\ⅆ}^5}{\ⅆx^5}y\left(x\right)\right)-45\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)\left(\frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)\right)\left(\frac{{\ⅆ}^4}{\ⅆx^4}y\left(x\right)\right)+40\frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)$

$\mathrm{liesol}\left(\mathrm{ode}\,y\left(x\right)\right)$

$\left\{y\left(x\right)=\int \left(\int \mathrm{RootOf}\left(-\left({\int }_{\phantom{\mathrm{`\; `}}}^{\mathrm{\_Z}}\frac{1}{\mathrm{RootOf}\left(-20\ln \left(\mathrm{\_f}\right)+{\int }_{\phantom{\mathrm{`\; `}}}^{\mathrm{\_Z}}\mathrm{\_k}\left({\ⅇ}^{\mathrm{RootOf}\left(81{\mathrm{\_k}}^2{\ⅇ}^{\mathrm{\_Z}}+20{\ⅇ}^{\mathrm{\_Z}}\ln \left({\ⅇ}^{\mathrm{\_Z}}+27\right)-40{\ⅇ}^{\mathrm{\_Z}}\ln \left(2\right)-20{\ⅇ}^{\mathrm{\_Z}}\ln \left(5\right)+162\mathrm{c\_\_1}{\ⅇ}^{\mathrm{\_Z}}-20\mathrm{\_Z}{\ⅇ}^{\mathrm{\_Z}}+2187{\mathrm{\_k}}^2+540\ln \left({\ⅇ}^{\mathrm{\_Z}}+27\right)-1080\ln \left(2\right)-540\ln \left(5\right)+4374\mathrm{c\_\_1}-540\mathrm{\_Z}-540\right)}+27\right)\ⅆ\mathrm{\_k}+20\mathrm{c\_\_2}\right)}\ⅆ\mathrm{\_f}\right)+x+\mathrm{c\_\_3}\right)\ⅆx\right)\ⅆx+\mathrm{c\_\_4}x+\mathrm{c\_\_5}\right\}$