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 integrate_sols
 integrate the solutions of a differential operator or equation

 Calling Sequence integrate_sols(L, domain) integrate_sols(eqn, dvar)

Parameters

 L - differential operator domain - list containing two names eqn - homogeneous linear differential equation dvar - dependent variable

Description

 • The input L is a differential operator. This procedure computes an operator M of minimal order such that any solution of L has an antiderivative which is a solution of M.
 • If the order of L equals the order of M then the output is a list [M, r] such that r(f) is an antiderivative of $f$ and also a solution of M for every solution $f$ of L. If the order of L is not equal to M then only M is given in the output. In this case M equals $L\mathrm{Dt}$ where $\mathrm{Dt}$ is the derivation.
 • The argument domain describes the differential algebra. If this argument is the list $\left[\mathrm{Dt},t\right]$, then the differential operators are notated with the symbols $\mathrm{Dt}$ and $t$. They are viewed as elements of the differential algebra $C\left(t\right)$ $\left[\mathrm{Dt}\right]$ where $C$ is the field of constants.
 • If the argument domain is omitted then the differential specified by the environment variable _Envdiffopdomain is used. If this environment variable is not set then the argument domain may not be omitted.
 • Instead of a differential operator, the input can also be a linear homogeneous differential equation having rational function coefficients. In this case the second argument must be the dependent variable.
 • This function is part of the DEtools package, and so it can be used in the form integrate_sols(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[integrate_sols](..).

Examples

 > $\mathrm{with}\left(\mathrm{DEtools}\right):$
 > $L≔\mathrm{Dz}+\frac{1}{z}$
 ${L}{≔}{\mathrm{Dz}}{+}\frac{{1}}{{z}}$ (1)
 > $\mathrm{integrate_sols}\left(L,\left[\mathrm{Dz},z\right]\right)$
 $\left({\mathrm{Dz}}{+}\frac{{1}}{{z}}\right){}{\mathrm{Dz}}$ (2)

The result was only the operator M. For an example where both M and r are obtained, consider:

 > $L≔{\mathrm{Dz}}^{2}-\frac{3\mathrm{Dz}}{2z\left(z-1\right)}-\frac{\left(z-2\right)\left({z}^{2}-z+1\right)}{z{\left(z-1\right)}^{2}}$
 ${L}{≔}{{\mathrm{Dz}}}^{{2}}{-}\frac{{3}}{{2}}{}\frac{{\mathrm{Dz}}}{{z}{}\left({z}{-}{1}\right)}{-}\frac{\left({z}{-}{2}\right){}\left({{z}}^{{2}}{-}{z}{+}{1}\right)}{{z}{}{\left({z}{-}{1}\right)}^{{2}}}$ (3)
 > $\mathrm{integrate_sols}\left(L,\left[\mathrm{Dz},z\right]\right)$
 $\left[{-}\frac{{z}{}{{\mathrm{Dz}}}^{{2}}}{{z}{-}{1}}{+}\frac{{1}}{{2}}{}\frac{{\mathrm{Dz}}}{{\left({z}{-}{1}\right)}^{{2}}}{+}{1}{,}\frac{{z}{}{\mathrm{Dz}}}{{z}{-}{1}}{-}\frac{{1}}{{2}{}{\left({z}{-}{1}\right)}^{{2}}}\right]$ (4)

Regarding the meaning of the second element in the output of integrate_sols, consider the following second order ODE and its solution.

 > $\mathrm{ode}≔\left(27{x}^{2}+4\right)\left(\frac{ⅆ}{ⅆx}\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)\right)+27x\left(\frac{ⅆ}{ⅆx}y\left(x\right)\right)-3y\left(x\right)$
 ${\mathrm{ode}}{≔}\left({27}{}{{x}}^{{2}}{+}{4}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}{}{y}{}\left({x}\right)\right){+}{27}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)\right){-}{3}{}{y}{}\left({x}\right)$ (5)
 > $\mathrm{sol}≔\mathrm{dsolve}\left(\mathrm{ode},y\left(x\right)\right)$
 ${\mathrm{sol}}{≔}{y}{}\left({x}\right){=}{\mathrm{_C1}}{}{\mathrm{sin}}{}\left(\frac{{1}}{{3}}{}{\mathrm{arctan}}{}\left(\frac{{3}{}\sqrt{{3}}{}{x}}{\sqrt{{-}{27}{}{{x}}^{{2}}{-}{4}}}\right)\right){+}{\mathrm{_C2}}{}{\mathrm{cos}}{}\left(\frac{{1}}{{3}}{}{\mathrm{arctan}}{}\left(\frac{{3}{}\sqrt{{3}}{}{x}}{\sqrt{{-}{27}{}{{x}}^{{2}}{-}{4}}}\right)\right)$ (6)

The ODE satisfied by the integral of sol is given by the first element of the output of integrate_sols while the formula for computing its solution (as a function of sol) is given by the second element.

 > $\mathrm{int_s}≔\mathrm{integrate_sols}\left(\mathrm{ode},y\left(x\right)\right):$
 > $\mathrm{ode2}≔{\mathrm{int_s}}_{1}$
 ${\mathrm{ode2}}{≔}{y}{}\left({x}\right){-}\frac{{9}}{{8}}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)\right){+}\left(\frac{{9}}{{8}}{}{{x}}^{{2}}{+}\frac{{1}}{{6}}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}{}{y}{}\left({x}\right)\right)$ (7)
 > $\mathrm{integral_of_sol}≔{\mathrm{int_s}}_{2}$
 ${\mathrm{integral_of_sol}}{≔}\frac{{9}}{{8}}{}{x}{}{y}{}\left({x}\right){+}\left({-}\frac{{9}}{{8}}{}{{x}}^{{2}}{-}\frac{{1}}{{6}}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)\right)$ (8)

So the solution to ode2 is obtained by substituting sol into integral_of_sol.

 > $\mathrm{sol2}≔y\left(x\right)=\mathrm{simplify}\left(\mathrm{eval}\left(\mathrm{integral_of_sol},\mathrm{sol}\right)\right)$
 ${\mathrm{sol2}}{≔}{y}{}\left({x}\right){=}\frac{{1}}{{24}}{}\frac{{27}{}{\mathrm{sin}}{}\left(\frac{{1}}{{3}}{}{\mathrm{arctan}}{}\left(\frac{{3}{}\sqrt{{3}}{}{x}}{\sqrt{{-}{27}{}{{x}}^{{2}}{-}{4}}}\right)\right){}\sqrt{{3}}{}{\mathrm{_C2}}{}{{x}}^{{2}}{-}{27}{}{\mathrm{cos}}{}\left(\frac{{1}}{{3}}{}{\mathrm{arctan}}{}\left(\frac{{3}{}\sqrt{{3}}{}{x}}{\sqrt{{-}{27}{}{{x}}^{{2}}{-}{4}}}\right)\right){}\sqrt{{3}}{}{\mathrm{_C1}}{}{{x}}^{{2}}{+}{27}{}{\mathrm{sin}}{}\left(\frac{{1}}{{3}}{}{\mathrm{arctan}}{}\left(\frac{{3}{}\sqrt{{3}}{}{x}}{\sqrt{{-}{27}{}{{x}}^{{2}}{-}{4}}}\right)\right){}\sqrt{{-}{27}{}{{x}}^{{2}}{-}{4}}{}{\mathrm{_C1}}{}{x}{+}{27}{}{\mathrm{cos}}{}\left(\frac{{1}}{{3}}{}{\mathrm{arctan}}{}\left(\frac{{3}{}\sqrt{{3}}{}{x}}{\sqrt{{-}{27}{}{{x}}^{{2}}{-}{4}}}\right)\right){}\sqrt{{-}{27}{}{{x}}^{{2}}{-}{4}}{}{\mathrm{_C2}}{}{x}{+}{4}{}{\mathrm{sin}}{}\left(\frac{{1}}{{3}}{}{\mathrm{arctan}}{}\left(\frac{{3}{}\sqrt{{3}}{}{x}}{\sqrt{{-}{27}{}{{x}}^{{2}}{-}{4}}}\right)\right){}\sqrt{{3}}{}{\mathrm{_C2}}{-}{4}{}{\mathrm{cos}}{}\left(\frac{{1}}{{3}}{}{\mathrm{arctan}}{}\left(\frac{{3}{}\sqrt{{3}}{}{x}}{\sqrt{{-}{27}{}{{x}}^{{2}}{-}{4}}}\right)\right){}\sqrt{{3}}{}{\mathrm{_C1}}}{\sqrt{{-}{27}{}{{x}}^{{2}}{-}{4}}}$ (9)
 > $\mathrm{odetest}\left(\mathrm{sol2},\mathrm{ode2}\right)$
 ${0}$ (10)

That sol2 is the integral of sol can also be verified as follows: Differentiate sol2 and verify that it is equal to sol.

 > $\mathrm{simplify}\left(\frac{\partial }{\partial x}\mathrm{rhs}\left(\mathrm{sol2}\right)-\mathrm{rhs}\left(\mathrm{sol}\right)\right)$
 ${0}$ (11)

References

 Abramov, S.A., and van Hoeij, M. "A method for the Integration of Solutions of Ore Equations." ISSAC '97 Proceedings, pp. 172-175. 1997.
 van der Put, M., and Singer, M. F. Galois Theory of Linear Differential Equations, Vol. 328. Springer, 2003. An electronic version of this book is available at http://www4.ncsu.edu/~singer/ms_papers.html.

 See Also