Given a differential equation (ODE) to solve. The first step is to obtain the general solution, which will have a number of arbitrary constants in it depending on order of the ode. The next step is to use the supplied initial and/or boundary conditions to solve for the arbitrary constants in the general solution.
Typically for first order ode, we need one IC to solve for the constant. But more can be given, including a mixture of initial and boundary conditions. As long as it is possible to resolve the arbitrary constant using all given IC’s, there is no restriction on how many IC one can give.
For an example, for the ode \(y^{\prime }+y=\sin \left ( x\right ) \), the solution is \(y=-\frac {\cos x}{2}+\frac {\sin \left ( x\right ) }{2}+c_{1}e^{-x}\). If the initial condition was \(y\left ( 0\right ) =0\) then this gives \(c_{1}=\frac {1}{2}\) and the particular solution becomes \(y=-\frac {\cos x}{2}+\frac {\sin \left ( x\right ) }{2}+\frac {e^{-x}}{2}\). If the IC was given as
We see that now \(c_{1}=\frac {1+\cos \left ( 1\right ) -\sin \left ( 1\right ) }{2\left ( 1+\cosh \left ( 1\right ) -\sinh \left ( 3\right ) \right ) }\). We can also supply more than one IC. If the IC was \(y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =0\) the solution will become
We see this is the same solution when the IC was \(y\left ( 0\right ) =0\).
This note is to talk about the algorithm to use to solve for the constants of integrations that will work for all cases and regardless of how many are given. As long as the IC is satisfied, it is not important they be minimal number or all of the same type. Mixed IC are also allowed.