4.4.2.6 Example 6
Let solve the same ode above but with now this IC \(y^{\prime }\left ( 0\right ) =1\).
\begin{align*} yy^{\prime \prime }+y^{\prime }\left ( y^{\prime }-1\right ) & =0\\ y^{\prime }\left ( 0\right ) & =1 \end{align*}
From the above, we found that 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 \(x_{0},y_{0}^{\prime }\) only but we are not given \(y_{0}\). Because of this, we will use (3) and not
(4) and use the values for the given \(x_{0},y_{0}^{\prime }\) where needed and replace \(y_{0}\) by \(y\left ( 0\right ) \). Hence (3)
becomes
\begin{align*} \int _{x_{0}}^{x}a_{0}\left ( \alpha ,y,y^{\prime }\right ) d\alpha +\int _{y_{\left ( 0\right ) }}^{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\\ 0+\int _{0}^{y}\left ( y^{\prime }-1\right ) d\beta +\int _{y_{0}^{\prime }}^{y^{\prime }}y_{0}d\gamma & =0\\ 0+\left ( y^{\prime }-1\right ) \int _{y\left ( 0\right ) }^{y}d\beta +y_{0}\int _{1}^{y^{\prime }}\left ( 0\right ) d\gamma & =0\\ \left ( y^{\prime }-1\right ) \left ( y-y\left ( 0\right ) \right ) & =0 \end{align*}
Hence \(y=-y\left ( 0\right ) \) or \(\left ( y^{\prime }-1\right ) =0\) which gives solution and \(y=x+c\). But \(y=-y\left ( 0\right ) \) does not satisfies the IC \(y^{\prime }\left ( 0\right ) =1\). But \(y=x+c\) does. Hence the
solution is
\[ y=x+c \]