One of the significant improvements for solving linear ODEs is the method which makes use of the factorization of differential operators (see diffop, and examples,diffop). As an example, the third order linear ODE

ode := x^2*diff(y(x),x$3) + (-x^3+3*x)*diff(y(x),x$2) + (-x^3+3*x)*y(x);

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

is solved as

dsolve(ode);

$y\left(x\right)=\mathrm{c\_\_1}\mathrm{BesselJ}\left(1\,x\right)+\mathrm{c\_\_2}\mathrm{BesselY}\left(1\,x\right)+\mathrm{c\_\_3}\left(-\mathrm{BesselJ}\left(1\,x\right)\int \frac{\mathrm{BesselY}\left(1\,x\right){\ⅇ}^{\frac{x^2}{2}}}{x}\ⅆx+\int \frac{\mathrm{BesselJ}\left(1\,x\right){\ⅇ}^{\frac{x^2}{2}}}{x}\ⅆx\mathrm{BesselY}\left(1\,x\right)\right)$

The method of solution takes advantage of the (sometimes) possible factorization of the linear differential operator associated to the given LODE. For example, for ode above, the differential operator is given by

DEtools[de2diffop](ode, y(x), [Dx,x]);

$x^2{\mathrm{Dx}}^3+\left(-x^3+3x\right){\mathrm{Dx}}^2-x^3+3x$

and this operator is factored as

DEtools[DFactor]((3), [Dx,x]);

$\left[x^2\mathrm{Dx}-x\left(x^2-2\right)\,{\mathrm{Dx}}^2+\frac{\mathrm{Dx}}{x}+\frac{x^2-1}{x^2}\right]$

The original ode is then solved by solving the LODEs determined by these two factors above (the solution to the left factor enters as a non-homogeneous term in the LODE associated to the right factor; solving this non-homogeneous LODE leads to the solution of ode).

As the second example, consider

ode := diff(y(x),x$3) - x*diff(y(x),x$2) - b^2*diff(y(x),x) + b^2*x*y(x);

$\mathrm{ode}≔\frac{{\ⅆ}^3}{\ⅆx^3}y\left(x\right)-x\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)-b^2\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+b^{2}xy\left(x\right)$

This LODE is solved by dsolve as follows

dsolve(ode);

$y\left(x\right)=\mathrm{c\_\_1}{\ⅇ}^{bx}+\mathrm{c\_\_2}{\ⅇ}^{-bx}+\mathrm{c\_\_3}\left(\mathrm{erf}\left(\frac{\mathrm{I}}{2}\sqrt{2}\left(b-x\right)\right){\ⅇ}^{2bx}+\mathrm{erf}\left(\frac{\mathrm{I}}{2}\sqrt{2}\left(b+x\right)\right)\right){\ⅇ}^{-\frac{b\left(b+2x\right)}{2}}$

In this case the solving method is taking advantage of the partial factorization

Dx^3-x*Dx^2-b^2*Dx+b^2*x = (Dx-x)*(Dx^2-b^2);

${\mathrm{Dx}}^3-x{\mathrm{Dx}}^2-b^2\mathrm{Dx}+b^{2}x=\left(\mathrm{Dx}-x\right)\left({\mathrm{Dx}}^2-b^2\right)$

rather than the complete factorization

Dx^3-x*Dx^2-b^2*Dx+b^2*x = (Dx-x)*(Dx-b)*(Dx+b);

${\mathrm{Dx}}^3-x{\mathrm{Dx}}^2-b^2\mathrm{Dx}+b^{2}x=\left(\mathrm{Dx}-x\right)\left(\mathrm{Dx}-b\right)\left(\mathrm{Dx}+b\right)$

Even in cases where dsolve cannot find solutions for all the factors, it can still return a reduction of order expressed using a DESol structure. For example, for

ode := diff(y(x),x$3) + (x^3-x)*diff(y(x),x$2) - (x^4+3)*diff(y(x),x) - (x^3-x)*y(x);

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

we have

dsolve(ode);

$y\left(x\right)=\mathrm{c\_\_1}{\ⅇ}^{\frac{x^2}{2}}+\mathrm{c\_\_2}\int {\ⅇ}^{-\frac{x^2\left(x^2+2\right)}{4}}\mathrm{HeunB}\left(-\frac{1}{2}\,0\,-\frac{3}{2}\,1\,\frac{x^2}{2}\right)\ⅆx{\ⅇ}^{\frac{x^2}{2}}+\mathrm{c\_\_3}\int x{\ⅇ}^{-\frac{x^2\left(x^2+2\right)}{4}}\mathrm{HeunB}\left(\frac{1}{2}\,0\,-\frac{3}{2}\,1\,\frac{x^2}{2}\right)\ⅆx{\ⅇ}^{\frac{x^2}{2}}$

that is, a reduction of order by one, since, in the factorization

DEtools[de2diffop](ode, y(x), [Dx,x]);

${\mathrm{Dx}}^3+\left(x^3-x\right){\mathrm{Dx}}^2+\left(-x^4-3\right)\mathrm{Dx}-x^3+x$

DEtools[DFactor]((11), [Dx,x]);

$\left[{\mathrm{Dx}}^2+x^3\mathrm{Dx}-1\,\mathrm{Dx}-x\right]$

dsolve only succeeded in solving the right factor (the left factor has no Liouvillian solutions).