diffeq*diffeq - Maple Help
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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
 hadamardproduct
 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)

 See Also