The exactsol routine determines whether the first argument is an exact first order ODE and, if so, returns a solution to the equation. When the equation is not exact, the routine tries to find an integrating factor that converts the equation into an equivalent exact 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 exactsol(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[exactsol](..).

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

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

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

$\left\{z\left(t\right)=\frac{\mathrm{c\_\_1}}{t}\,z\left(t\right)=\frac{-\frac{\mathrm{c\_\_1}}{2}-\frac{\mathrm{I}\sqrt{3}\mathrm{c\_\_1}}{2}}{t}\,z\left(t\right)=\frac{-\frac{\mathrm{c\_\_1}}{2}+\frac{\mathrm{I}\sqrt{3}\mathrm{c\_\_1}}{2}}{t}\right\}$

$\mathrm{ode}≔\left(3t^4{z\left(t\right)}^2-t^2\right)\mathrm{D}\left(z\right)\left(t\right)+\left(4t^3{z\left(t\right)}^3-2tz\left(t\right)\right)=0\:$

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

$\left\{z\left(t\right)=\frac{\frac{{\left(\left(12\sqrt{3}\sqrt{27{\mathrm{c\_\_1}}^2-4t^2}-108\mathrm{c\_\_1}\right)t^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{6t}+\frac{2t}{{\left(\left(12\sqrt{3}\sqrt{27{\mathrm{c\_\_1}}^2-4t^2}-108\mathrm{c\_\_1}\right)t^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{t}\,z\left(t\right)=\frac{-\frac{{\left(\left(12\sqrt{3}\sqrt{27{\mathrm{c\_\_1}}^2-4t^2}-108\mathrm{c\_\_1}\right)t^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{12t}-\frac{t}{{\left(\left(12\sqrt{3}\sqrt{27{\mathrm{c\_\_1}}^2-4t^2}-108\mathrm{c\_\_1}\right)t^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-\frac{\mathrm{I}\sqrt{3}\left(\frac{{\left(\left(12\sqrt{3}\sqrt{27{\mathrm{c\_\_1}}^2-4t^2}-108\mathrm{c\_\_1}\right)t^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{6t}-\frac{2t}{{\left(\left(12\sqrt{3}\sqrt{27{\mathrm{c\_\_1}}^2-4t^2}-108\mathrm{c\_\_1}\right)t^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}{2}}{t}\,z\left(t\right)=\frac{-\frac{{\left(\left(12\sqrt{3}\sqrt{27{\mathrm{c\_\_1}}^2-4t^2}-108\mathrm{c\_\_1}\right)t^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{12t}-\frac{t}{{\left(\left(12\sqrt{3}\sqrt{27{\mathrm{c\_\_1}}^2-4t^2}-108\mathrm{c\_\_1}\right)t^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+\frac{\mathrm{I}\sqrt{3}\left(\frac{{\left(\left(12\sqrt{3}\sqrt{27{\mathrm{c\_\_1}}^2-4t^2}-108\mathrm{c\_\_1}\right)t^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{6t}-\frac{2t}{{\left(\left(12\sqrt{3}\sqrt{27{\mathrm{c\_\_1}}^2-4t^2}-108\mathrm{c\_\_1}\right)t^2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}{2}}{t}\right\}$