The linearsol routine determines whether the first argument is a first order linear ODE and, if so, returns a solution to the equation.

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 linearsol(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[linearsol](..).

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

$\mathrm{ode}≔\mathrm{diff}\left(z\left(t\right)\,t\right)+p\left(t\right)z\left(t\right)=q\left(t\right)$

$\mathrm{ode}≔\frac{\ⅆ}{\ⅆt}z\left(t\right)+p\left(t\right)z\left(t\right)=q\left(t\right)$

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

$\left\{z\left(t\right)=\left(\int q\left(t\right){\ⅇ}^{\int p\left(t\right)\ⅆt}\ⅆt+\mathrm{\_C1}\right){\ⅇ}^{\int -p\left(t\right)\ⅆt}\right\}$

$\mathrm{ode}≔\mathrm{D}\left(z\right)\left(t\right)+t^{2}z\left(t\right)=1-t+t^2$

$\mathrm{ode}≔\mathrm{D}\left(z\right)\left(t\right)+t^{2}z\left(t\right)=t^2-t+1$

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

$\left\{z\left(t\right)=\left(-1+{\ⅇ}^{\frac{t^3}{3}}+\frac{3^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(\mathrm{-1}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(\frac{t^2{\left(\mathrm{-1}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\mathrm{\Gamma }\left(\frac{2}{3}\right)}{{\left(-t^3\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-\frac{t^2{\left(\mathrm{-1}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\mathrm{\Gamma }\left(\frac{2}{3}\,-\frac{t^3}{3}\right)}{{\left(-t^3\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}{3}-\frac{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\left(\mathrm{-1}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(\frac{2t\sqrt{3}{\left(\mathrm{-1}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\mathrm{\pi }}{3\mathrm{\Gamma }\left(\frac{2}{3}\right){\left(-t^3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-\frac{t{\left(\mathrm{-1}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\mathrm{\Gamma }\left(\frac{1}{3}\,-\frac{t^3}{3}\right)}{{\left(-t^3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}{3}+\mathrm{\_C1}\right){\ⅇ}^{-\frac{t^3}{3}}\right\}$