2.4 Quadrature ode

\begin{align*} y^{\prime } & =f\left ( x\right ) \\ y^{\prime } & =f\left ( y\right ) \end{align*}

The following flow chart gives the algorithm for solving quadrature ode.  Internal code is first_order_ode_quadrature

Figure 2.3: Flow chart for first order quadrature

Solved by direct integration. There are two forms. They are

\begin{align*} y^{\prime } & =f\left ( x\right ) \\ y^{\prime } & =f\left ( y\right ) \end{align*}

For first form, the solution is

\[ y=\int f\left ( x\right ) dx+c \]

For the second form the solution is

\begin{align*} \int \frac {dy}{f\left ( y\right ) } & =\int dx\qquad f\left ( y\right ) \neq 0\\ \int \frac {dy}{f\left ( y\right ) } & =x+c \end{align*}

These two forms are special cases of separable first order ode \(y^{\prime }=f\left ( x\right ) g\left ( y\right ) .\)

For the form \(y^{\prime }=f\left ( y\right ) \) and if IC are given, we should always check if IC satisfies the ODE itself first. If so, then the solution is simply \(y=y_{0}\). i.e. there is no need to integrate and solve for constant of integration and any of this. This only works for \(y^{\prime }=f\left ( y\right ) \) form. Not for \(y^{\prime }=f\left ( x\right ) \).

Given an ode \(y^{\prime }=f\left ( x\right ) \) and if it is not possible to integrate \(\int f\left ( x\right ) dx\), then the final solution should be left as

\[ y\left ( x\right ) =\int f\left ( x\right ) dx+c_{1}\]

If initial conditions are given as \(y\left ( x_{0}\right ) =y_{0}\) then the above is adjusted to become

\[ y\left ( x\right ) =\int _{x_{0}}^{x}f\left ( \tau \right ) d\tau +y_{0}\]

This is only when the integration of \(f\left ( x\right ) \) can not be computed.

On the other hand, if the ode is \(y^{\prime }=g\left ( y\right ) \) and it is also not possible to integrate \(\int \frac {1}{g\left ( y\right ) }\) then the final answer now becomes

\[ \int ^{y\left ( x\right ) }\frac {1}{g\left ( \tau \right ) }d\tau =x+c_{1}\]

If initial conditions are given as \(y\left ( x_{0}\right ) =y_{0}\) then the above is adjusted to become

\[ \int _{0}^{y\left ( x\right ) }\frac {1}{g\left ( \tau \right ) }d\tau +\int _{0}^{y_{0}}\frac {1}{g\left ( \tau \right ) }d\tau =x-x_{0}\]

Or

\[ y\left ( x\right ) =\operatorname {RootOf}\left ( \int _{\_Z}^{y_{0}}\frac {1}{g\left ( \tau \right ) }d\tau +x-x_{0}\right ) \]

For the case where it is not possible to solve for \(y^{\prime }\) explicitly, then \(\operatorname {RootOf}\) is used. For example, given

\[ \sin \left ( y^{\prime }\right ) +y^{\prime }=x \]

This is quadrature, since it has only \(y^{\prime }\) and \(x\). But it is not possible to isolate \(y^{\prime }\). The solution will be in terms of \(\operatorname {RootOf}\) given by

\[ y^{\prime }=\operatorname {RootOf}\left ( \sin \left ( \_Z\right ) +\_Z-x\right ) \]

We now still continue as before and integrate both sides which results in

\[ y\left ( x\right ) =\int \operatorname {RootOf}\left ( \sin \left ( \_Z\right ) +\_Z-x\right ) dx+c \]

If initial conditions are given as \(y\left ( x_{0}\right ) =y_{0}\) the above is modified to become

\[ y\left ( x\right ) -y_{0}=\int _{x_{0}}^{x}\operatorname {RootOf}\left ( \sin \left ( \_Z\right ) +\_Z-\tau \right ) d\tau \]

What happens if the ode had a missing \(x\) instead? For an example

\[ \sin \left ( y^{\prime }\right ) +y^{\prime }=y \]

Now solving for \(y^{\prime }\) gives

\[ y^{\prime }=\operatorname {RootOf}\left ( \sin \left ( \_Z\right ) +\_Z-y\right ) \]

Integrating as before results in

\begin{align*} \frac {dy}{\operatorname {RootOf}\left ( \sin \left ( \_Z\right ) +\_Z-y\right ) } & =dx\\ \int \frac {dy}{\operatorname {RootOf}\left ( \sin \left ( \_Z\right ) +\_Z-y\right ) } & =\int dx\\ \int _{0}^{y\left ( x\right ) }\frac {d\tau }{\operatorname {RootOf}\left ( \sin \left ( \_Z\right ) +\_Z-\tau \right ) } & =x+c \end{align*}

If initial conditions \(y\left ( x_{0}\right ) =y_{0}\) are given, the above becomes

\[ \int _{0}^{y_{0}}\frac {d\tau }{\operatorname {RootOf}\left ( \sin \left ( \_Z\right ) +\_Z-\tau \right ) }+\int _{y_{0}}^{y\left ( x\right ) }\frac {d\tau }{\operatorname {RootOf}\left ( \sin \left ( \_Z\right ) +\_Z-\tau \right ) }=x-x_{0}\]