4.9.1.1.11 Example 11 \(y^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}+y^{\prime }=0,y^{\prime }\left ( 1\right ) =3\)

\begin{align} y^{\prime \prime }+\left ( y^{\prime }\right ) ^{2}+y^{\prime } & =0\tag {1}\\ y^{\prime }\left ( 1\right ) & =3\nonumber \end{align}

Let \(p=y^{\prime }\) then \(y^{\prime \prime }=\frac {dp}{dx}=\frac {dp}{dy}\frac {dy}{dx}=\frac {dp}{dy}p\). Hence the ode becomes

\begin{align} \frac {dp}{dy}p+p^{2}+p & =0\tag {2}\\ p^{\prime }p+p^{2}+p & =0\nonumber \\ p^{\prime }+p+1 & =0\hspace {0.5in}p\neq 0\nonumber \end{align}

This is quadrature.

\begin{align*} \frac {dp}{dy}\frac {1}{p+1} & =-1\\ \int dp\frac {1}{p+1} & =-\int dy\\ \ln \left ( p+1\right ) & =-y+c_{1}\\ p+1 & =c_{1}e^{-y}\end{align*}

But \(p=y^{\prime }\), hence the above becomes

\begin{align} y^{\prime }+1 & =c_{1}e^{-y}\nonumber \\ y^{\prime } & =c_{1}e^{-y}-1\nonumber \\ \frac {dy}{dx}\frac {1}{c_{1}e^{-y}-1} & =1\nonumber \\ \int \frac {dy}{c_{1}e^{-y}-1} & =\int dx\nonumber \\ -\ln \left ( e^{y}-c_{1}\right ) & =x+c_{2}\nonumber \\ \ln \left ( e^{y}-c_{1}\right ) & =-x+c_{3}\nonumber \\ e^{y}-c_{1} & =c_{4}e^{-x}\nonumber \\ e^{y} & =c_{4}e^{-x}+c_{1}\nonumber \\ y & =\ln \left ( c_{4}e^{-x}+c_{1}\right ) \tag {2}\end{align}

At \(x=1\) we have \(y^{\prime }\left ( 1\right ) =3\). From the above solution we see that \(y^{\prime }=\frac {\frac {d}{dx}\left ( c_{4}e^{-x}+c_{1}\right ) }{c_{4}e^{-x}+c_{1}}=\frac {-c_{4}e^{-x}}{c_{4}e^{-x}+c_{1}}\). At \(x=1\)

\begin{align*} 3 & =\frac {-c_{4}e^{-1}}{c_{4}e^{-1}+c_{1}}\\ 3c_{4}e^{-1}+3c_{1} & =-c_{4}e^{-1}\\ 4c_{4}e^{-1}+3c_{1} & =0\\ c_{1} & =-\frac {4}{3}c_{4}e^{-1}\end{align*}

Hence (2) becomes

\[ y=\ln \left ( c_{4}e^{-x}-\frac {4}{3}c_{4}e^{-1}\right ) \]