DEtools - Maple Help

Home : Support : Online Help : Mathematics : Differential Equations : Lie Symmetry Method : Commands for ODEs : DEtools/ode_int_y

DEtools

 ode_int_y
 given the nth order linear ODE satisfied by y(x), compute the nth order linear ODE satisfied by int(y(x),x)
 ode_y1
 given the nth order linear ODE satisfied by y(x), compute the nth order linear ODE satisfied by diff(y(x),x)

 Calling Sequence ode_int_y(ode, y(x)) ode_y1(ode, y(x))

Parameters

 ode - ordinary differential equation satisfied by y(x) y(x) - unknown function of one variable

Description

 • Given a nth order linear ODE for $y\left(x\right)$, the ode_int_y and ode_y1 commands respectively compute the nth order linear ODE satisfied by $∫y\left(x\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}ⅆx$ and $\frac{ⅆ}{ⅆx}y\left(x\right)$.

Examples

For enhanced input output use DEtools[diff_table] and PDEtools[declare].

 > $\mathrm{with}\left(\mathrm{DEtools},\mathrm{diff_table},\mathrm{ode_int_y},\mathrm{ode_y1}\right)$
 $\left[{\mathrm{diff_table}}{,}{\mathrm{ode_int_y}}{,}{\mathrm{ode_y1}}\right]$ (1)
 > $\mathrm{PDEtools}[\mathrm{declare}]\left(\mathrm{prime}=x,y\left(x\right),c\left(x\right)\right)$
 ${\mathrm{derivatives with respect to}}{}{x}{}{\mathrm{of functions of one variable will now be displayed with \text{'}}}$
 ${y}{}\left({x}\right){}{\mathrm{will now be displayed as}}{}{y}$
 ${c}{}\left({x}\right){}{\mathrm{will now be displayed as}}{}{c}$ (2)
 > $Y≔\mathrm{diff_table}\left(y\left(x\right)\right):$
 > $\mathrm{PDEtools}[\mathrm{declare}]\left(y\left(x\right),c\left(x\right),\mathrm{prime}=x\right)$
 ${y}{}\left({x}\right){}{\mathrm{will now be displayed as}}{}{y}$
 ${c}{}\left({x}\right){}{\mathrm{will now be displayed as}}{}{c}$
 ${\mathrm{derivatives with respect to}}{}{x}{}{\mathrm{of functions of one variable will now be displayed with \text{'}}}$ (3)

Now, if $y\left(x\right)$ satisfies

 > $c[0]\left(x\right){Y}_{[]}+c[1]\left(x\right){Y}_{x}+c[2]\left(x\right){Y}_{x,x}+{Y}_{x,x,x,x}=0$
 ${y}{}{{c}}_{{0}}{+}{\mathrm{y\text{'}}}{}{{c}}_{{1}}{+}{\mathrm{y\text{'}\text{'}}}{}{{c}}_{{2}}{+}{\mathrm{y\text{'}\text{'}\text{'}\text{'}}}{=}{0}$ (4)

then the derivative of $y\left(x\right)$ satisfies

 > $\mathrm{DEtools}[\mathrm{ode_y1}]\left(\right)=0$
 ${\mathrm{y\text{'}\text{'}\text{'}\text{'}}}{-}\frac{{{c}}_{{0}}^{{\mathrm{\text{'}}}}{}{\mathrm{y\text{'}\text{'}\text{'}}}}{{{c}}_{{0}}}{+}{{c}}_{{2}}{}{\mathrm{y\text{'}\text{'}}}{-}\frac{\left({{c}}_{{0}}^{{\mathrm{\text{'}}}}{}{{c}}_{{2}}{-}{{c}}_{{1}}{}{{c}}_{{0}}{-}{{c}}_{{2}}^{{\mathrm{\text{'}}}}{}{{c}}_{{0}}\right){}{\mathrm{y\text{'}}}}{{{c}}_{{0}}}{-}\frac{\left({{c}}_{{0}}^{{\mathrm{\text{'}}}}{}{{c}}_{{1}}{-}{{c}}_{{0}}^{{2}}{-}{{c}}_{{1}}^{{\mathrm{\text{'}}}}{}{{c}}_{{0}}\right){}{y}}{{{c}}_{{0}}}{=}{0}$ (5)

and so, the integral of the function $y\left(x\right)$ in the equation above satisfy this other ODE (the starting point)

 > $\mathrm{DEtools}[\mathrm{ode_int_y}]\left(,y\left(x\right)\right)=0$
 ${y}{}{{c}}_{{0}}{+}{\mathrm{y\text{'}}}{}{{c}}_{{1}}{+}{\mathrm{y\text{'}\text{'}}}{}{{c}}_{{2}}{+}{\mathrm{y\text{'}\text{'}\text{'}\text{'}}}{=}{0}$ (6)