The basic task being performed by dsolve when solving an "Initial Conditions" (ICs) ODE problem is to find appropriate values for the set of integration constants _Cn appearing in the symbolic solution of the problem, such that the solution will match the given ICs.

As general rules for IC problems, the first argument must be a set containing an ODE or a system of ODEs together with the ICs, the second argument must be a set containing the indeterminate functions of the problem, and the number of ICs should not be greater than the sum of the differential orders of the given ODEs (see PDEtools[difforder]).

If no variable is specified, $x$ is assumed to be the variable.

For symbolic problems (that is, when neither series nor numeric solutions were requested) a typical IC can be any equation relating algebraic expressions, or just the algebraic expressions themselves, then assumed to be = 0 (see examples below).

The derivatives entering the ICs can always be expressed using the D syntax (for example $\mathrm{D}\left(y\right)\left(0\right)=1$, $\mathrm{D}\left(y\right)\left(A+B\right)=C$). Alternately, standard math syntax may be used in 2-D math (for example $y\'\left(0\right)=1$ is equivalent to the first example above, and $y^{\left(2\right)}\left(x\right)$ is equivalent to $\mathrm{D}\left(\mathrm{D}\left(y\left(x\right)\right)\right)$ or ${\mathrm{D}}^{\left(2\right)}\left(y\left(x\right)\right)$. If the evaluation points are of type symbol, diff will also work (for example, $\frac{\ⅆ}{\ⅆa}y\left(a\right)=1$ means that the derivative of $y$ at $a$ is 1).

It is also possible to give "coupled" ICs, involving more than one function in each IC equation, and perhaps in a nonlinear manner. When nonlinear ICs are given dsolve might return a sequence of solution sets related to the various possible solutions found for the integration constants.

When requesting numeric or series solutions, by giving the extra argument 'type=numeric' or 'type=series'; see dsolve,numeric, or dsolve,series), or the use of integral transforms (see dsolve,inttrans), the ICs must be given as equations. All derivatives entering the ICs must be expressed using the D syntax, each IC must be related to a single indeterminate function (coupled ICs are not allowed), and all ICs must be linear in the indeterminate function or its derivatives.

$\mathrm{ode}≔\mathrm{diff}\left(y\left(t\right)\,t\,t\right)+{\mathrm{diff}\left(y\left(t\right)\,t\right)}^2=0$

$\mathrm{ode}≔\frac{{\ⅆ}^2}{\ⅆt^2}y\left(t\right)+{\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)}^2=0$

$\mathrm{ans4}\left[1\right]≔\mathrm{dsolve}\left(\left\{\mathrm{ode}\,y\left(0\right)=3\right\}\,y\left(t\right)\right)$

${\mathrm{ans4}}_1≔y\left(t\right)=\ln \left(\mathrm{c\_\_1}t+{\ⅇ}^3\right)$

$\mathrm{ans4}\left[2\right]≔\mathrm{dsolve}\left(\left\{\mathrm{ode}\,y\left(0\right)=3\,\mathrm{D}\left(y\right)\left(0\right)=0\right\}\,y\left(t\right)\right)$

${\mathrm{ans4}}_2≔y\left(t\right)=3$

$\mathrm{ans4}\left[3\right]≔\mathrm{dsolve}\left(\left\{\mathrm{ode}\,\mathrm{diff}\left(y\left(a\right)\,a\,a\right)=A\right\}\,y\left(t\right)\right)$

${\mathrm{ans4}}_3≔y\left(t\right)=\ln \left(\mathrm{c\_\_1}t-\frac{\left(a\sqrt{-A}-1\right)\mathrm{c\_\_1}}{\sqrt{-A}}\right),y\left(t\right)=\ln \left(\mathrm{c\_\_1}t-\frac{\left(a\sqrt{-A}+1\right)\mathrm{c\_\_1}}{\sqrt{-A}}\right)$

$\mathrm{ans4}\left[4\right]≔\mathrm{dsolve}\left(\left\{\mathrm{ode}\,{\mathrm{D}}^{\left(2\right)}\left(y\right)\left(a\right)-y\left(b\right)=A\,\exp \left(y\left(b\right)\right)=B\right\}\,y\left(t\right)\right)$

${\mathrm{ans4}}_4≔y\left(t\right)=\ln \left(t\mathrm{RootOf}\left(\left(A{a}^2-2Aab+A{b}^2+a^2\ln \left(B\right)-2ab\ln \left(B\right)+b^2\ln \left(B\right)+1\right){\mathrm{\_Z}}^2+\ln \left(B\right)+A+\left(2Aa-2Ab+2\ln \left(B\right)a-2\ln \left(B\right)b\right)\mathrm{\_Z}\right)B-b\mathrm{RootOf}\left(\left(A{a}^2-2Aab+A{b}^2+a^2\ln \left(B\right)-2ab\ln \left(B\right)+b^2\ln \left(B\right)+1\right){\mathrm{\_Z}}^2+\ln \left(B\right)+A+\left(2Aa-2Ab+2\ln \left(B\right)a-2\ln \left(B\right)b\right)\mathrm{\_Z}\right)B+B\right)$

$\mathrm{map}\left(\mathrm{odetest}\,\left[\mathrm{ans4}\left[1\right]\,\mathrm{ans4}\left[2\right]\,\mathrm{ans4}\left[3\right]\,\mathrm{ans4}\left[4\right]\right]\,\mathrm{ode}\right)$

$\left[0\,0\,0\,0\,0\right]$

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

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

$\mathrm{IC\_1}≔\left\{x\left(a\right)=A\,y\left(b\right)=B\right\}$

$\mathrm{IC\_1}≔\left\{x\left(a\right)=A\,y\left(b\right)=B\right\}$

$\mathrm{ans1}≔\mathrm{combine}\left(\mathrm{dsolve}\left(\mathrm{sys}\mathbf{union}\mathrm{IC\_1}\,\left\{x\left(t\right)\,y\left(t\right)\right\}\right)\,\mathrm{trig}\right)$

$\mathrm{ans1}≔\left\{x\left(t\right)=\frac{A\cos \left(b-t\right)-B\sin \left(a-t\right)}{\cos \left(a-b\right)}\,y\left(t\right)=\frac{A\sin \left(b-t\right)+B\cos \left(a-t\right)}{\cos \left(a-b\right)}\right\}$

$\mathrm{IC\_2}≔\left\{\mathrm{diff}\left(x\left(a\right)\,a\right)=B\,x\left(a\right)=A\right\}$

$\mathrm{IC\_2}≔\left\{\frac{\ⅆ}{\ⅆa}x\left(a\right)=B\,x\left(a\right)=A\right\}$

$\mathrm{ans2}≔\mathrm{combine}\left(\mathrm{dsolve}\left(\mathrm{sys}\mathbf{union}\mathrm{IC\_2}\,\left\{x\left(t\right)\,y\left(t\right)\right\}\right)\,\mathrm{trig}\right)$

$\mathrm{ans2}≔\left\{x\left(t\right)=A\cos \left(a-t\right)-B\sin \left(a-t\right)\,y\left(t\right)=A\sin \left(a-t\right)+B\cos \left(a-t\right)\right\}$

$\mathrm{IC\_3}≔\left\{\mathrm{diff}\left(x\left(a\right)\,a\right)=A\,\mathrm{diff}\left(y\left(b\right)\,b\right)=B\right\}$

$\mathrm{IC\_3}≔\left\{\frac{\ⅆ}{\ⅆa}x\left(a\right)=A\,\frac{\ⅆ}{\ⅆb}y\left(b\right)=B\right\}$

$\mathrm{ans3}≔\mathrm{combine}\left(\mathrm{dsolve}\left(\mathrm{sys}\mathbf{union}\mathrm{IC\_3}\,\left\{x\left(t\right)\,y\left(t\right)\right\}\right)\,\mathrm{trig}\right)$

$\mathrm{ans3}≔\left\{x\left(t\right)=-\frac{A\sin \left(b-t\right)+B\cos \left(a-t\right)}{\cos \left(a-b\right)}\,y\left(t\right)=\frac{A\cos \left(b-t\right)-B\sin \left(a-t\right)}{\cos \left(a-b\right)}\right\}$

$\mathrm{map}\left(\mathrm{odetest}\,\left[\mathrm{ans1}\,\mathrm{ans2}\,\mathrm{ans3}\right]\,\mathrm{sys}\right)$

$\left[\left\{0\right\}\,\left\{0\right\}\,\left\{0\right\}\right]$