The eulersols routine returns a basis of the space of solutions of a linear differential equation of Euler type (also sometimes called Cauchy or Legendre type). These are equations of the form

There are two input forms. The first has as the first argument a linear differential equation in diff or D form and as the second argument the variable in the differential equation.

A second input sequence accepts for the first argument the list of coefficients of a linear ode, and for the second argument the independent variable of the lode. This input sequence is convenient for programming with the eulersols routine.

In the second calling sequence, the list of coefficients is given in order from low differential order to high differential order and does not include the nonhomogeneous term.

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

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

$\mathrm{ode}≔t^2\mathrm{diff}\left(z\left(t\right)\,\mathrm{`\$`}\left(t\,2\right)\right)-t\mathrm{diff}\left(z\left(t\right)\,t\right)-3z\left(t\right)=0\:$

$\mathrm{eulersols}\left(\mathrm{ode}\,z\left(t\right)\right)$

$\left[t^3\,\frac{1}{t}\right]$

$\mathrm{ode}≔t^3{\mathrm{D}}^{\left(3\right)}\left(z\right)\left(t\right)+3t^2{\mathrm{D}}^{\left(2\right)}\left(z\right)\left(t\right)-2t\mathrm{D}\left(z\right)\left(t\right)+2z\left(t\right)=0$

$\mathrm{ode}≔t^3{\mathrm{D}}^{\left(3\right)}\left(z\right)\left(t\right)+3t^2{\mathrm{D}}^{\left(2\right)}\left(z\right)\left(t\right)-2t\mathrm{D}\left(z\right)\left(t\right)+2z\left(t\right)=0$

$\mathrm{eulersols}\left(\mathrm{ode}\,z\left(t\right)\right)$

$\left[\frac{1}{t^2}\,t\,t\ln \left(t\right)\right]$

$\mathrm{ode}≔\left(t^2-2t+1\right)\mathrm{diff}\left(z\left(t\right)\,\mathrm{`\$`}\left(t\,2\right)\right)-\left(t-1\right)\mathrm{diff}\left(z\left(t\right)\,t\right)-3z\left(t\right)=0\:$

$\mathrm{eulersols}\left(\mathrm{ode}\,z\left(t\right)\right)$

$\left[{\left(t-1\right)}^3\,\frac{1}{t-1}\right]$

$\mathrm{eulersols}\left(\left[-3\,1-t\,t^2-2t+1\right]\,t\right)$

$\left[{\left(t-1\right)}^3\,\frac{1}{t-1}\right]$

$\mathrm{ode}≔3t^5{\mathrm{D}}^{\left(5\right)}\left(z\right)\left(t\right)+t^2{\mathrm{D}}^{\left(2\right)}\left(z\right)\left(t\right)-t\mathrm{D}\left(z\right)\left(t\right)-3z\left(t\right)=0$

$\mathrm{ode}≔3t^5{\mathrm{D}}^{\left(5\right)}\left(z\right)\left(t\right)+t^2{\mathrm{D}}^{\left(2\right)}\left(z\right)\left(t\right)-t\mathrm{D}\left(z\right)\left(t\right)-3z\left(t\right)=0$

$\mathrm{eulersols}\left(\mathrm{ode}\,z\left(t\right)\right)$

$\left[t^3\,t^{\mathrm{RootOf}\left(3{\mathrm{\_Z}}^4-21{\mathrm{\_Z}}^3+42{\mathrm{\_Z}}^2-23\mathrm{\_Z}+1\,\mathrm{index}=1\right)}\,t^{\mathrm{RootOf}\left(3{\mathrm{\_Z}}^4-21{\mathrm{\_Z}}^3+42{\mathrm{\_Z}}^2-23\mathrm{\_Z}+1\,\mathrm{index}=2\right)}\,t^{\mathrm{RootOf}\left(3{\mathrm{\_Z}}^4-21{\mathrm{\_Z}}^3+42{\mathrm{\_Z}}^2-23\mathrm{\_Z}+1\,\mathrm{index}=3\right)}\,t^{\mathrm{RootOf}\left(3{\mathrm{\_Z}}^4-21{\mathrm{\_Z}}^3+42{\mathrm{\_Z}}^2-23\mathrm{\_Z}+1\,\mathrm{index}=4\right)}\right]$