2.5.1.1 Solved using first_order_ode_dAlembert

Entering first order ode dAlembert solver

\[\begin {aligned} x {y^{\prime }}^{2}-y+2 y^{\prime }&=0 \end {aligned}\]

Let \(p=y^{\prime }\) the ode becomes

\begin{align*} x \,p^{2}+2 p -y = 0 \end{align*}

Solving for \(y\) from the above results in

\begin{align*} y &= x \,p^{2}+2 p\tag {1A} \end{align*}

This has the form

\begin{align*} y&=x f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= p^{2}\\ g &= 2 p \end{align*}

Hence (2) becomes

\begin{align*} -p^{2}+p = \left (2 x p +2\right ) p^{\prime }\left (x \right )\tag {3} \end{align*}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} -p^{2}+p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0\\ p_{2} &=1 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y&=0\\ y&=x +2 \end{align*}

Inverting Eq (2A) gives

\begin{align*} \frac {d}{d p}x \left (p \right ) = \frac {2 x \left (p \right ) p +2}{-p^{2}+p}\tag {4} \end{align*}

This ODE is now solved for \(x \left (p \right )\). The integrating factor is

\begin{align*} \mu &= {\mathrm e}^{\int \frac {2}{p -1}d p}\\ \mu &= \left (p -1\right )^{2}\\ \mu &= \left (p -1\right )^{2}\tag {5} \end{align*}

Integrating gives

\begin{align*} x \left (p \right )&= \frac {1}{\mu } \left ( \int { \mu \left (-\frac {2}{p \left (p -1\right )}\right ) \,dp} + c_1\right )\\ &= \frac {1}{\mu } \left (c_1 +\frac {c_1 -2 p +2 \ln \left (p \right )}{\left (p -1\right )^{2}}\right ) \\ &= \frac {c_1 -2 p +2 \ln \left (p \right )}{\left (p -1\right )^{2}}\tag {5} \end{align*}

Now we need to eliminate \(p\) between the above solution and \(y = x \,p^{2}+2 p\). The first method is to solve for \(p\) from \(y = x \,p^{2}+2 p\) and substitute the result into Eq. (5). The Second method is to solve for \(p\) from Eq. (5) and substitute the result into \(y = x \,p^{2}+2 p\).

Eliminating \(p\) from the following two equations

\begin{gather*} \begin {aligned} x &= \frac {c_1 -2 p +2 \ln \left (p \right )}{\left (p -1\right )^{2}} \\ y &= x \,p^{2}+2 p \end {aligned} \end{gather*}

results in

\begin{gather*} \begin {aligned} p &= {\mathrm e}^{\operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )} \\ \end {aligned} \end{gather*}

Substituting the above into \(y = x \,p^{2}+2 p\) and simplifying gives

\begin{gather*} \begin {aligned} y &= x \,{\mathrm e}^{2 \operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )}+2 \,{\mathrm e}^{\operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )} \\ \end {aligned} \end{gather*}

Simplifying \(y = x \,{\mathrm e}^{2 \operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )}+2 \,{\mathrm e}^{\operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )}\) gives

\[ y = 2 x \,{\mathrm e}^{\operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )}+2 \operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )+c_1 -x \]

Isoclines	Solutions plot

Summary of solutions found

\begin{align*} y&=0\\ y&=x +2\\ y&=2 x \,{\mathrm e}^{\operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )}+2 \operatorname {RootOf}\left (-x \,{\mathrm e}^{2 \textit {\_Z}}+2 x \,{\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}}+c_1 +2 \textit {\_Z} -x \right )+c_1 -x \end{align*}