4.4.2.5 Example 5

Let solve the same ode above but with only one IC is given and not both. In other words, if we are given either \(y\left ( x_{0}\right ) =y_{0}\) or \(y^{\prime }\left ( x_{0}\right ) =y_{0}^{\prime }\) only. To see how to handle this method in such case. We know if there are no IC are given, then we use EQ (4) above, which is

\begin{equation} \int _{0}^{x}a_{0}\left ( \alpha ,y,y^{\prime }\right ) d\alpha +\int _{0}^{y}a_{1}\left ( 0,\beta ,y^{\prime }\right ) d\beta +\int _{0}^{y^{\prime }}a_{2}\left ( 0,0,\gamma \right ) d\gamma =c_{1} \tag {4}\end{equation}

And if both initial conditions are given, then we use EQ (3), which is

\begin{equation} \int _{x_{0}}^{x}a_{0}\left ( \alpha ,y,y^{\prime }\right ) d\alpha +\int _{y_{0}}^{y}a_{1}\left ( x_{0},\beta ,y^{\prime }\right ) d\beta +\int _{y_{0}^{\prime }}^{y^{\prime }}a_{2}\left ( x_{0},y_{0},\gamma \right ) d\gamma =0 \tag {3}\end{equation}

Let see what to do when only one IC is given for the second order ode

\begin{align*} yy^{\prime \prime }+y^{\prime }\left ( y^{\prime }-1\right ) & =0\\ y\left ( 0\right ) & =0 \end{align*}

From problem 4, we found that this match works

\begin{align} a_{2} & =y\tag {2A}\\ a_{1} & =\left ( y^{\prime }-1\right ) \nonumber \\ a_{0} & =0\nonumber \end{align}

And now we are given \(x_{0},y_{0}\) only but we are not given \(y_{0}^{\prime }\). Because of this, we will use (3) and not (4) and use the values for the given \(x_{0},y_{0}\) where needed and replace \(y_{0}^{\prime }\) by \(y^{\prime }\left ( 0\right ) \). Hence (3) becomes

\begin{align*} \int _{x_{0}}^{x}a_{0}\left ( \alpha ,y,y^{\prime }\right ) d\alpha +\int _{y_{0}}^{y}a_{1}\left ( x_{0},\beta ,y^{\prime }\right ) d\beta +\int _{y^{\prime }\left ( 0\right ) }^{y^{\prime }}a_{2}\left ( x_{0},y_{0},\gamma \right ) d\gamma & =0\\ 0+\int _{0}^{y}\left ( y^{\prime }-1\right ) d\beta +\int _{y^{\prime }\left ( 0\right ) }^{y^{\prime }}y_{0}d\gamma & =0\\ 0+\left ( y^{\prime }-1\right ) y+\int _{y^{\prime }\left ( 0\right ) }^{y^{\prime }}\left ( 0\right ) d\gamma & =0\\ \left ( y^{\prime }-1\right ) y & =0 \end{align*}

Hence \(y=0\) or \(y^{\prime }=1\). But \(y^{\prime }=1\). Solving this gives \(y=x+c_{1}\). using initial conditions \(y\left ( 0\right ) =0\) gives \(c=0\). Hence \(y=x\) is also a solution. Hence solutions are

\begin{align*} y & =0\\ y & =x \end{align*}

This shows that if we are given even partial initial conditions, then we should use EQ (3) and not EQ (4). The following example gives one more illustration of this.