diffeq*diffeq - Maple Help

gfun

 diffeq+diffeq
 determine the differential equation satisfied by the sum of two holonomic functions
 diffeq*diffeq
 determine the differential equation satisfied by the Cauchy product of two holonomic functions
 determine the differential equation satisfied by the Hadamard product of two holonomic functions

 Calling Sequence diffeq+diffeq(eq1, eq2, y(z)) diffeq*diffeq(eq1, eq2, y(z)) hadamardproduct(eq1, eq2, y(z))

Parameters

 eq1, eq2 - two linear differential equations with polynomial coefficients y - name; holonomic function name z - name; variable of the holonomic function y

Description

 • The gfun[diffeq+diffeq](eq1, eq2, y(z)) command determines the differential equation satisfied by the sum of two holonomic functions, eq1 and eq2.
 If f and g are holonomic function solutions of eq1 and eq2, respectively, the gfun[diffeq+diffeq] function returns a linear differential equation verified by $f+g$.
 The differential order of the returned equation is at most the sum of eq1 and eq2's differential orders for the gfun[diffeq+diffeq] function.
 • The gfun[diffeq*diffeq](eq1, eq2, y(z)) command determines the differential equation satisfied by the Cauchy product of two holonomic functions, eq1 and eq2.
 If f and g are holonomic function solutions of eq1 and eq2), respectively, the gfun[diffeq*diffeq] function returns a linear differential equation verified by $fg$.
 The differential order of the returned equation is at most the product of eq1 and eq2's differential orders for the gfun[diffeq*diffeq] function.
 • The gfun[hadamardproduct](eq1, eq2, y(z)) command determines the differential equation satisfied by the Hadamard product of two holonomic functions, eq1 and eq2.
 If f and g are holonomic function solutions of eq1 and eq2, respectively, the gfun[hadamardproduct] function returns a linear differential equation verified by the Hadamard product of f and g.  The function whose coefficient of ${z}^{n}$ in the Taylor expansion around 0 is the product of the corresponding coefficients of f and g.

Examples

 > $\mathrm{with}\left(\mathrm{gfun}\right):$
 > $\mathrm{eq1}≔\mathrm{D}\left(y\right)\left(x\right)-y\left(x\right):$
 > $\mathrm{eq2}≔\left(1+x\right){\mathrm{D}}^{\left(2\right)}\left(y\right)\left(x\right)+\mathrm{D}\left(y\right)\left(x\right):$
 > $\mathrm{diffeq+diffeq}\left(\mathrm{eq1},\mathrm{eq2},y\left(x\right)\right)$
 $\left({-}{x}{-}{3}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left({-}{{x}}^{{2}}{-}{2}{}{x}{+}{1}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left({{x}}^{{2}}{+}{3}{}{x}{+}{2}\right){}\left(\frac{{{ⅆ}}^{{3}}}{{ⅆ}{{x}}^{{3}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)$ (1)
 > $\mathrm{diffeq*diffeq}\left(\mathrm{eq1},\mathrm{eq2},y\left(x\right)\right)$
 ${x}{}{y}{}\left({x}\right){+}\left({-}{2}{}{x}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left({1}{+}{x}\right){}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)$ (2)
 > $\mathrm{hadamardproduct}\left(\mathrm{eq1},\mathrm{eq2},y\left(x\right)\right)$
 $\left\{\left({1}{+}{x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){+}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}{-}{{\mathrm{_t}}}_{{2}}{}{{\mathrm{_C}}}_{{1}}{,}{y}{}\left({0}\right){=}{{\mathrm{_t}}}_{{2}}{}{{\mathrm{_C}}}_{{0}}\right\}$ (3)