The dsolve command with the numeric or type=numeric option and an initial value problem (IVP) finds a numerical solution for the ODE or ODE system IVP. If the optional equation method=numericmethod is provided (where numericmethod is one of rkf45, ck45, rosenbrock, dverk78, lsode, gear, taylorseries, or classical), dsolve uses that method to obtain the numerical solution.

The return value of dsolve and the following high level or common options are discussed in dsolve[numeric] and dsolve[Error_Control].

The exception is the list input form for abserr which requires the discussion below on conversion to first order to understand fully.

In most cases it is sufficient to specify a single value for abserr for all solution components. The exception to this is when some dependent variables for derivatives of the system have values on a different scale than other variables of the system. When this occurs, the list input form for abserr can be used to specify a different absolute error tolerance for each dependent variable of the system, or each solution component of the system (the difference between these is that specification of a tolerance for a dependent variable of the system also does so for derivatives of that variable, while for solution components each value is considered independently.

In order to specify dependent or component abserr, the order of the dependent variables in the output must be known. This can be controlled by specifying the dependent variables of the problem (the vars parameter to dsolve) as a list.

From this, dependent variable abserr is easy to understand as there is a 1-1 correspondence between the specified variables and the specified abserr.

The per-component abserr is a little trickier as the basic idea is the same, but depending on the order of each variable more than one abserr value may be required.

For example, for a problem containing $x\text{'}\text{'},y\text{'}\text{'}\text{'},z\'$, if the variables were specified as $\left[x\left(t\right)\,y\left(t\right)\,z\left(t\right)\right]$, then for per-component abserr six values would be required, and would correspond to $\[x\left(t\right),x\'\left(t\right),y\left(t\right),y\'\left(t\right),y\text{'}\text{'}\left(t\right),z\left(t\right)\]$.

Only the rkf45, ck45, and rosenbrock IVP methods support the list form for abserr.

The default IVP method is a Runge-Kutta Fehlberg method, which produces a solution accurate to fifth order. Another non-stiff IVP method is a Runge-Kutta method with the Cash-Karp coefficients. The default stiff method is a Rosenbrock method, which produces a solution accurate to fourth order. See dsolve[rkf45], dsolve[ck45], and dsolve[rosenbrock] for more detail on these methods. The other available methods for IVP are dsolve[dverk78], dsolve[lsode], dsolve[taylorseries], dsolve[gear], and dsolve[classical].

All IVP methods, with the exception of taylorseries, can work with a complex-valued IVP with a real-valued independent variable.

The following options are specific to IVP, and are concerned with specification of the input system in procedure form (procopts):

and the following options, also specific to IVP (and in some cases DAE), allow control of the solution method and step size, and the use of parameters in the system:

Before discussion of the procedure-based options, the Maple treatment of the numerical handling of ODE system IVPs is described. To begin with, all systems are converted to first order systems by the introduction of new dependent variables. For example, if the single ODE

were to be numerically solved, the new dependent variables $u=y\'$, $v=y\text{'}\text{'}$ would be introduced, and the single third order ODE would become a system of three first order ODEs in three dependent variables. In addition, the derivatives of the first order system are isolated. The result would be

and the initial conditions would transform in the same way.

Now that the ODE is converted to a first order system, it is possible to describe the procedure-based options more precisely. In all that follows, any reference to the system or dependent variables refers only to the first order system.

$\mathrm{dsys1}≔\left\{\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+y\left(x\right)=0\,y\left(1\right)=\sin \left(1\right)\,\mathrm{D}\left(y\right)\left(1\right)=\cos \left(1\right)\right\}$

$\mathrm{dsys1}≔\left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+y\left(x\right)=0\,y\left(1\right)=\sin \left(1\right)\,\mathrm{D}\left(y\right)\left(1\right)=\cos \left(1\right)\right\}$

$\mathrm{dsol1}≔\mathrm{dsolve}\left(\mathrm{dsys1}\,\mathrm{numeric}\,\mathrm{method}=\mathrm{gear}\right)$

$\mathrm{dsol1}≔\mathbf{proc}\left(\mathrm{x\_gear}\right)...\mathbf{end\; proc}$

$\mathrm{dsol1}\left(0\right)$

$\left[x=0.\,y\left(x\right)=8.05339006126137\times {10}^{\mathrm{-12}}\,\frac{\ⅆ}{\ⅆx}y\left(x\right)=0.999999999997424\right]$

$\mathrm{dsol1}\left(\mathrm{\pi }\right)$

$\left[x=3.14159265358979\,y\left(x\right)=6.97710003877356\times {10}^{\mathrm{-12}}\,\frac{\ⅆ}{\ⅆx}y\left(x\right)=\mathrm{-0.999999999999881}\right]$

$\mathrm{deq2}≔\left\{\mathrm{diff}\left(x\left(t\right)\,t\right)=y\left(t\right)\,\mathrm{diff}\left(y\left(t\right)\,t\right)=x\left(t\right)+y\left(t\right)\right\}$

$\mathrm{deq2}≔\left\{\frac{\ⅆ}{\ⅆt}x\left(t\right)=y\left(t\right)\,\frac{\ⅆ}{\ⅆt}y\left(t\right)=x\left(t\right)+y\left(t\right)\right\}$

$\mathrm{ic2}≔\left\{x\left(0\right)=2\,y\left(0\right)=1\right\}\:$

$\mathrm{dsol2}≔\mathrm{dsolve}\left(\mathrm{deq2}\mathbf{union}\mathrm{ic2}\,\mathrm{numeric}\,\mathrm{output}=\mathrm{listprocedure}\,\mathrm{range}=0..1\right)$

$\mathrm{dsol2}≔\left[t=\mathbf{proc}\left(t\right)...\mathbf{end\; proc}\,x\left(t\right)=\mathbf{proc}\left(t\right)...\mathbf{end\; proc}\,y\left(t\right)=\mathbf{proc}\left(t\right)...\mathbf{end\; proc}\right]$

$\mathrm{dsol2x}≔\mathrm{subs}\left(\mathrm{dsol2}\,x\left(t\right)\right)\:$$\mathrm{dsol2y}≔\mathrm{subs}\left(\mathrm{dsol2}\,y\left(t\right)\right)\:$

$\left[\mathrm{dsol2x}\left(0\right)\,\mathrm{dsol2y}\left(0\right)\right]$

$\left[2.\,1.\right]$

$\left[\mathrm{dsol2x}\left(0.4\right)\,\mathrm{dsol2y}\left(0.4\right)\right]$

$\left[2.69118458493332\,2.60811748317261\right]$

$\left[\mathrm{dsol2x}\left(1.0\right)\,\mathrm{dsol2y}\left(1.0\right)\right]$

$\left[5.58216753140384\,7.82688926499912\right]$

dproc3 := proc(N,t,Y,YP)
   YP[1] := sigma * (Y[2] - Y[1]);
   YP[2] := r*Y[1] - Y[1]*Y[3] - Y[2];
   YP[3] := Y[1]*Y[2] - b*Y[3];
end proc:

$\mathrm{\sigma }≔10\:$$r≔28\:$$b≔\frac{8}{3}\:$

$\mathrm{ic3}≔\mathrm{Array}\left(\left[5\,5\,5\right]\right)\:$

$\mathrm{dvars3}≔\left[x\left(t\right)\,y\left(t\right)\,z\left(t\right)\right]\:$

$\mathrm{dsol3}≔\mathrm{dsolve}\left(\mathrm{numeric}\,\mathrm{number}=3\,\mathrm{procedure}=\mathrm{dproc3}\,\mathrm{start}=0\,\mathrm{initial}=\mathrm{ic3}\,\mathrm{procvars}=\mathrm{dvars3}\right)$

$\mathrm{dsol3}≔\mathbf{proc}\left(\mathrm{x\_rkf45}\right)...\mathbf{end\; proc}$

$\mathrm{dsol3}\left(1\right)$

$\left[t=1.\,x\left(t\right)=\mathrm{-7.09064281000129}\,y\left(t\right)=\mathrm{-4.13869080943593}\,z\left(t\right)=29.0616065518357\right]$

$\mathrm{dsol3}\left(2\right)$

$\left[t=2.\,x\left(t\right)=\mathrm{-9.13349830723850}\,y\left(t\right)=\mathrm{-12.6955376022149}\,z\left(t\right)=22.4637998042670\right]$

$\mathrm{dsol3}\left(3\right)$

$\left[t=3.\,x\left(t\right)=\mathrm{-5.00807269946467}\,y\left(t\right)=\mathrm{-2.53517706788747}\,z\left(t\right)=26.6155378375388\right]$

dproc4 := proc(N,x,Y,YP)
   YP[1] := exp(-x)*Y[2];
   YP[2] := -sin(x)*Y[1];
end proc;

$\mathrm{dproc4}≔\mathbf{proc}\left(N\,x\,Y\,\mathrm{YP}\right)\mathrm{YP}\[1\]≔\exp \left(\−x\right)\*Y\[2\]\;\mathrm{YP}\[2\]≔\−\sin \left(x\right)\*Y\[1\]\mathbf{end\; proc}$

$\mathrm{ic4}≔\mathrm{Array}\left(\left[1\,0\right]\right)\:$

$\mathrm{dvars4}≔\left[y\left(x\right)\,\exp \left(x\right)\mathrm{diff}\left(y\left(x\right)\,x\right)\right]\:$

$\mathrm{dsol4}≔\mathrm{dsolve}\left(\mathrm{numeric}\,\mathrm{number}=2\,\mathrm{procedure}=\mathrm{dproc4}\,\mathrm{start}=0\,\mathrm{initial}=\mathrm{ic4}\,\mathrm{procvars}=\mathrm{dvars4}\right)$

$\mathrm{dsol4}≔\mathbf{proc}\left(\mathrm{x\_rkf45}\right)...\mathbf{end\; proc}$

$\mathrm{dsol4}\left(0\right)$

$\left[x=0.\,y\left(x\right)=1.\,{\ⅇ}^x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)=0.\right]$

$\mathrm{dsol4}\left(1\right)$

$\left[x=1.\,y\left(x\right)=0.924554126889498\,{\ⅇ}^x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)=\mathrm{-0.444194433465032}\right]$

$\mathrm{dsol4}\left(2\right)$

$\left[x=2.\,y\left(x\right)=0.741972301895190\,{\ⅇ}^x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)=\mathrm{-1.24013689804566}\right]$

$\mathrm{dsys1}≔\left\{\mathrm{diff}\left(y\left(x\right)\,x\right)+\mathrm{I}y\left(x\right)=0\,y\left(0\right)=1+2\mathrm{I}\right\}$

$\mathrm{dsys1}≔\left\{\frac{\ⅆ}{\ⅆx}y\left(x\right)+\mathrm{I}y\left(x\right)=0\,y\left(0\right)=1+2\mathrm{I}\right\}$

$\mathrm{dsol5}≔\mathrm{dsolve}\left(\mathrm{dsys1}\,\mathrm{numeric}\,\mathrm{method}=\mathrm{gear}\right)$

$\mathrm{dsol5}≔\mathbf{proc}\left(\mathrm{x\_gear}\right)...\mathbf{end\; proc}$

$\mathrm{dsol5}\left(\frac{\mathrm{\pi }}{2}\right)$

$\left[x=1.57079632679490\,y\left(x\right)=2.00000000000367-\mathrm{I}\right]$

$\mathrm{dsol5}\left(\mathrm{\pi }\right)$

$\left[x=3.14159265358979\,y\left(x\right)=\mathrm{-1.00000000001409}-2.00000000003328\mathrm{I}\right]$

$\mathrm{dsol5}\left(2\mathrm{\pi }\right)$

$\left[x=6.28318530717958\,y\left(x\right)=1.00000000000419+2.00000000002333\mathrm{I}\right]$

$\mathrm{dsys}≔\left\{\mathrm{diff}\left(y\left(x\right)\,x\right)={y\left(x\right)}^2\,y\left(0\right)=1\right\}$

$\mathrm{dsys}≔\left\{\frac{\ⅆ}{\ⅆx}y\left(x\right)={y\left(x\right)}^2\,y\left(0\right)=1\right\}$

$\mathrm{dsoli}≔\mathrm{dsolve}\left(\mathrm{dsys}\,\mathrm{numeric}\right)$

$\mathrm{dsoli}≔\mathbf{proc}\left(\mathrm{x\_rkf45}\right)...\mathbf{end\; proc}$

$\mathrm{dsoli}\left(0.5\right)$

$\left[x=0.5\,y\left(x\right)=2.00000045906252\right]$

$\mathrm{dsoli}\left('\mathrm{initial}'\right)$

$\left[x=0.\,y\left(x\right)=1.\right]$

$\mathrm{dsoli}\left('\mathrm{last}'\right)$

$\left[x=0.500000000000000\,y\left(x\right)=2.00000045906252\right]$

$\mathrm{dsoli}\left('\mathrm{method}'\right)$

$''rkf45''$

$\mathrm{dsoli}\left(1\right)$

Error, (in dsoli) cannot evaluate the solution further right of .99999999, probably a singularity

$\mathrm{dsoli}\left('\mathrm{last}'\right)$

$\left[x=0.999999994312848\,y\left(x\right)=1.00607191562369\times {10}^{12}\right]$

$\mathrm{dsoli}\left('\mathrm{initial}'=\left[1\,\frac{1}{2}\right]\right)$

$\left[x=1.\,y\left(x\right)=0.500000000000000\right]$

$\mathrm{dsoli}\left(3\right)$

Error, (in dsoli) cannot evaluate the solution further right of 2.9999999, probably a singularity

$\mathrm{dsoli}\left('\mathrm{last}'\right)$

$\left[x=2.99999998871727\,y\left(x\right)=3.25728571163544\times {10}^{11}\right]$

$\mathrm{dsoli2}≔\mathrm{dsolve}\left(\mathrm{dsys}\,\mathrm{numeric}\,\mathrm{output}=\mathrm{listprocedure}\right)$

$\mathrm{dsoli2}≔\left[x=\mathbf{proc}\left(x\right)...\mathbf{end\; proc}\,y\left(x\right)=\mathbf{proc}\left(x\right)...\mathbf{end\; proc}\right]$

$\mathrm{dsx}≔\mathrm{rhs}\left(\mathrm{dsoli2}\left[1\right]\right)$

$\mathrm{dsx}≔\mathbf{proc}\left(x\right)...\mathbf{end\; proc}$

$\mathrm{dsy}≔\mathrm{rhs}\left(\mathrm{dsoli2}\left[2\right]\right)$

$\mathrm{dsy}≔\mathbf{proc}\left(x\right)...\mathbf{end\; proc}$

$\mathrm{dsy}\left(0.5\right)$

$2.00000045906252$

$\mathrm{dsx}\left('\mathrm{last}'\right)$

$0.500000000000000$

$\mathrm{dsx}\left('\mathrm{initial}'=1\right)$

$1.$

$\mathrm{dsy}\left(1.5\right)$

$2.00000045906252$

$\mathrm{dsy}\left('\mathrm{initial}'=\left[1\,\frac{1}{2}\right]\right)$

$0.500000000000000$

$\mathrm{dsy}\left(3.0\right)$

Error, (in dsy) cannot evaluate the solution further right of 2.9999999, probably a singularity

$\left[\mathrm{dsx}\left('\mathrm{last}'\right)\,\mathrm{dsy}\left('\mathrm{last}'\right)\right]$

$\left[2.99999998871727\,3.25728571163544\times {10}^{11}\right]$